{"id":35814,"date":"2022-10-01T12:52:21","date_gmt":"2022-10-01T07:22:21","guid":{"rendered":"https:\/\/apboardsolutions.guru\/?p=35814"},"modified":"2022-10-01T12:52:21","modified_gmt":"2022-10-01T07:22:21","slug":"ap-inter-1st-year-physics-study-material-chapter-12","status":"publish","type":"post","link":"https:\/\/apboardsolutions.guru\/ap-inter-1st-year-physics-study-material-chapter-12\/","title":{"rendered":"AP Inter 1st Year Physics Study Material Chapter 12 Thermal Properties of Matter"},"content":{"rendered":"

Andhra Pradesh BIEAP\u00a0AP Inter 1st Year Physics Study Material<\/a> 12th Lesson Thermal Properties of Matter Textbook Questions and Answers.<\/p>\n

AP Inter 1st Year Physics Study Material 12th Lesson Thermal Properties of Matter<\/h2>\n

Very Short Answer Questions<\/span><\/p>\n

Question 1.
\nDistinguish between heat and temperature. [T.S. Mar. 15]
\nAnswer:
\nHeat<\/p>\n

    \n
  1. It is a form of energy.<\/li>\n
  2. It is the cause on the body.<\/li>\n
  3. It is determined with calorimeter.<\/li>\n
  4. It\u2019s unit is Joule and Calories.<\/li>\n<\/ol>\n

    Temperature<\/p>\n

      \n
    1. It is a degree of hotness or coldness of a body.<\/li>\n
    2. It is an effect on a body.<\/li>\n
    3. It is measured with thermometer.<\/li>\n
    4. It’s unit is degree’Celsius or Kelvin or degree Fahrenheit.<\/li>\n<\/ol>\n

      Question 2.
      \nWhat are the lower and upper fixing points in Celsius and Fahrenheit scales ? [T.S. Mar. 16]
      \nAnswer:
      \nIn Celsius scale, lower fixed point is ice point or 0\u00b0C and upper fixed point is steam point or 100\u00b0C. In Fahrenheit scale, lower fixed point is 32\u00b0F and upper fixed point is 212\u00b0F.<\/p>\n

      \"AP<\/p>\n

      Question 3.
      \nDo the values of coefficients of expansion differ, when the temperatures are measured on Centigrade scale or on Fahrenheit scale ?
      \nAnswer:
      \nYes. The coefficients of expansion depends on scale of temperature because
      \n\\(\\frac{\\alpha}{{ }^{\\circ} \\mathrm{C}}=\\frac{9}{5} \\alpha \/{ }^{\\circ} \\mathrm{F}\\)<\/p>\n

      Question 4.
      \nCan a substance contract on heating ? Give an example.
      \nAnswer:
      \nYes. Rubber, type metal, cast iron contract on heating.<\/p>\n

      Question 5.
      \nWhy gaps are left between rails on a railway track ? [A.P. Mar. 17, 16]
      \nAnswer:
      \nThe length of the rails increases in summer due to high temperature. Therefore a gap is left to allow this expansion.<\/p>\n

      Question 6.
      \nWhy do liquids have no linear and areal expansions ?
      \nAnswer:
      \nLiquids occupy the same shape of vessel. They do not have individual length and area. Hence, liquids have no linear and areal expansions.<\/p>\n

      Question 7.
      \nWhat is latent heat of fusion ?
      \nAnswer:
      \nThe amount of heat per unit mass required to change a substance from solid into liquid at the same temperature and pressure is called the latent heat of fusion (Lf<\/sub>).<\/p>\n

      \"AP<\/p>\n

      Question 8.
      \nWhat is latent heat of vapourisation ? [Mar. 13]
      \nAnswer:
      \nThe amount of heat per unit mass required to change a substance from the liquid to the vapour state at the same temperature and pressure is called the latent heat of vaporisation (Lv<\/sub>).<\/p>\n

      Question 9.
      \nWhat is specific gas constant ? Is it same for all gases ?
      \nAnswer:
      \nUniversal gas constant per molecular mass is called specific gas constant.
      \ni.e., r = \\(\\frac{\\mathrm{R}}{\\mathrm{m}}\\). No, it is different for different gases.<\/p>\n

      Question 10.
      \nWhat are the units and dimensions of specific gas constant ?
      \nAnswer:
      \nThe units of specific gas constant is J\/Kg\/K. Dimensional formula is (m0<\/sup>.L2<\/sup>t-2<\/sup>K1<\/sup>).<\/p>\n

      Question 11.
      \nWhy utensils are coated black ? Why the bottom of the utensils are made of copper ?
      \nAnswer:<\/p>\n

        \n
      1. Utensils are coated black, because it is a good absorber of heat.<\/li>\n
      2. Copper is a good conductor of heat. So, copper is used at the bottom of cooking utensils.<\/li>\n<\/ol>\n

        \"AP<\/p>\n

        Question 12.
        \nState Weins displacement law. [A.P. Mar. 17]
        \nAnswer:
        \nThe wavelength (\u03bbm<\/sub>) corresponding to maximum energy emitted by a black body is inversely proportional to its absolute temperature
        \ni.e., \u03bbm<\/sub> \u221d \\(\\frac{1}{T}\\)<\/p>\n

        Question 13.
        \nVentilators are provided in rooms just below the roof. Why ? [Mar. 14]
        \nAnswer:
        \nVentilators are provided in rooms just below the roof, because the hot air escapes out and fresh air enters into the rooms due to convection.<\/p>\n

        Question 14.
        \nDoes a body radiate heat at 0 K ? Does it radiate heat at 0\u00b0C ?
        \nAnswer:<\/p>\n

          \n
        1. No. A body does not radiate heat at 0k.<\/li>\n
        2. Yes. A body radiate heat at 0\u00b0C.<\/li>\n<\/ol>\n

          Question 15.
          \nState the different modes of transmission of heat. Which of these modes require medium ?
          \nAnswer:
          \nThe different modes of transmission are : 1) conduction 2) convection and 3) radiation.
          \nAmong three modes conduction and convection require medium.<\/p>\n

          Question 16.
          \nDefine coefficient of thermal conductivity and temperature gradient.
          \nAnswer:
          \nCoefficient of thermal conductivity : It is defined as the quantity of heat energy flowing per second between the opposite faces of cube of unit side, which are maintained at unit temperature difference.
          \nTemperature gradient: The change of temperature per unit length of conductor is called temperature gradient.<\/p>\n

          \"AP<\/p>\n

          Question 17.
          \nWhat is thermal resistance of a conductor ? On what factors does it depend ?
          \nAnswer:
          \nThe resistance offered by the conductor for the flow of heat is called thermal resistance (R) of the conductor. R = \\(\\frac{\\mathrm{d}}{\\mathrm{kA}}\\). It depends on<\/p>\n

            \n
          1. The nature of the material, i.e., thermal conductivity (K).<\/li>\n
          2. d; length of conductor along which heat flows and<\/li>\n
          3. A; area of cross section of the conductor.<\/li>\n<\/ol>\n

            Question 18.
            \nState the units and dimensions of coefficient of convection.
            \nAnswer:
            \nThe unit of coefficient of convection is, wm-2<\/sup>k-1<\/sup>.
            \nDimensional formula (m.L0<\/sup>T3<\/sup>\u03b8-1<\/sup>).<\/p>\n

            Question 19.
            \nDefine emissive power and emissivity ?
            \nAnswer:
            \nEmissive power : The energy flux emitted by unit surface area of a radiating body is known as emissive power (e\u03bb<\/sub>).<\/p>\n

            Missivity (e): It is defined as the ratio of the emissive power of the body to that of black body at the same temperature.<\/p>\n

            Question 20.
            \nWhat is greenhouse effect ? Explain global warming. [A.P. Mar. 15, 13]
            \nAnswer:
            \nGreenhouse effect: When the earth receives sun light, it gets heated up and emits infrared radiation into air. CO2<\/sub>, CH4<\/sub>, N2<\/sub>O, O3<\/sub>, Chlorofluoro carbon (green house gases) present in the air absorbs the heat content of infrared radiation and keeps the earth warm. This is called green house effect.<\/p>\n

            Global warming : As CO2<\/sub> content increases, more heat is retained in the atmosphere and the temperatures all over the world increases. This is called global warming.<\/p>\n

            Effects : a) Polar ice caps melt and fed more water in rivers and seas results in severe floods, b) In some areas, water resources dry up leading to drought conditions.<\/p>\n

            \"AP<\/p>\n

            Question 21.
            \nDefine absorptive power of a body. What is the absorptive power of a perfect black body ?
            \nAnswer:
            \nAbsorptive power : At a given temperature and wavelength, the ratio of the amount of radiant energy absorbed to the amount of radiant energy incident in a wavelength range is called the absorptive power at that temperature and wavelength. Absorptive power of a perfect black body is 1.<\/p>\n

            Question 22.
            \nState Newton’s law of cooling. [A.P. Mar. 18, 16; T.S. Mar. 18]
            \nAnswer:
            \nThe rate of loss of heat is directly proportional to the difference in temperature between the body and its surroundings provided the temperature difference is small. i.e., – \\(\\frac{\\mathrm{dQ}}{\\mathrm{dt}} \\alpha\\left(\\mathrm{T}_B-\\mathrm{T}_{\\mathrm{S}}\\right)\\)<\/p>\n

            Question 23.
            \nState the conditions under which Newton’s law of cooling is applicable.
            \nAnswer:<\/p>\n

              \n
            1. Loss of heat is negligible by conduction and only when it is due to convection.<\/li>\n
            2. Loss of heat occurs in a streamlined flow of air i.e., forced convection.<\/li>\n
            3. Temperature of the body is uniformly distributed over it.<\/li>\n<\/ol>\n

              Question 24.
              \nThe roof of buildings are often painted white during summer. Why ? [T.S. Mar. 17, 15]
              \nAnswer:
              \nThe roof of buildings are painted white during summer, because it reflects more heat radiations and absorbs less heat radiations. Hence we feel comfort inside the house.<\/p>\n

              Short Answer Questions<\/span><\/p>\n

              Question 1.
              \nExplain Celsius and Fahrenheit scales of temperature. Obtain the relation between Celsius and Fahrenheit scales of temperature.
              \nAnswer:
              \nCentigrade (Celsius) scale of temperature: In the Celsius scale of temperature, the lower fixed point is called the ice point and is assigned the value 0\u00b0C. The upper fixed point is called the steam point and is assigned the value 100\u00b0C. The interval between these two points (i.e., 100\u00b0C – 0 = 100\u00b0C) is subdivided into 100 equal parts each one corresponding to 1\u00b0C.
              \n\"AP
              \nFahrenheit scale of temperature : In the Fahrenheit scale of temperature, the lower fixed point is the ice point and is assigned a value 32\u00b0F and the upper fixed point is the staempoint and is assigned a value 212\u00b0F. The interval between these two points (i.e., 212\u00b0F – 32\u00b0F = 180\u00b0F) is subdivided into 180 equal parts, each one corresponding to 1\u00b0F.<\/p>\n

              Fahrenheit scale of temperature : In the Fahrenheit scale of temperature, the lower fixed point is the ice point and is assigned a value 32\u00b0F and the upper fixed point is the staempoint and is assigned a value 212\u00b0F. The interval between these two points (i.e., 212\u00b0F – 32\u00b0F = 180\u00b0F) is subdivided into 180 equal parts, each one corresponding to 1\u00b0F.
              \n\"AP
              \nRelation between Celsius and Fahreinheit scales of temperature:
              \nDifference of 100 Celsius degrees = Difference of 180 Fahrenheit degress
              \nWhen the temperature of a body is measured on both the Celsius and Fahrenheit scales, let the readings be tC<\/sub> and tF<\/sub> respectively. Then
              \n\\(\\frac{t_c-0}{t^{100}}=\\frac{t_F-32}{180} \\Rightarrow \\frac{t_C}{5}=\\frac{t_F-32}{9}\\)
              \nC = \\(\\frac{5}{9}\\) (F – 32)<\/p>\n

              \"AP<\/p>\n

              Question 2.
              \nTwo identical rectangular strips, one of copper and the other of steel, are riveted together to form a compound bar. What will happen on heating ?
              \nAnswer:
              \nBimetallic strip:
              \n\"AP
              \nTwo similar strips of copper and steel are revitted side by side as a compound strip called bimetallic strip. It is straight at ordinary temperatures. When the bimetallic strip is heated, copper expands more than steel. So, it bends with copper on convex side. When the bimetallic strip is cooled below room temperature, copper contracts more than steel and it bends with copper on concave side. Bimetallic strip is used in refrigerators, automatic iron, fire alarm etc.<\/p>\n

              Question 3.
              \nPendulum clocks generally go fast in winter and slow in summer. Why ?
              \nAnswer:
              \nThe time period of a pendulum at t1<\/sub>\u00b0C is T1<\/sub> = 2\u03c0 \\(\\sqrt{\\frac{L_1}{g}}\\) ………….. (1)
              \nWhere L1<\/sub> is length of pendulum at t1<\/sub>\u00b0C.
              \nIf L2<\/sub> is length of pendulum at t2<\/sub>\u00b0C,
              \nthen T2<\/sub> = 2\u03c0 \\(\\sqrt{\\frac{L_2}{g}}\\) ………………. (2)
              \n\\(\\frac{(2)}{(1)}\\) gives \\(\\frac{T_2}{T_1}=\\sqrt{\\frac{L_2}{L_1}}=\\sqrt{\\frac{L_1(1+\\alpha t)}{L_1}}\\)
              \nWhere a is the linear coefficient of expansion of pendulum clock. Where t = t2<\/sub> – t1<\/sub>
              \n\\(\\frac{T_2}{T_1}=\\sqrt{(1+\\alpha t)}=(1+\\alpha t)^{\\frac{1}{2}}\\)
              \n\\(\\frac{T_2}{T_1}=1+\\frac{\\alpha t}{2}\\)
              \n\\(\\frac{T_2}{T_1}-1=\\frac{\\alpha t}{2}\\)
              \n\\(\\frac{T_2-T_1}{T_1}=\\frac{\\alpha t}{2}\\)
              \n\\(\\frac{T_2-T_1}{T_1}\\) = time lost by pendulum per second.
              \nDue to expansion in length, during summer, time period increases or the clock looses time in summer. In winter due to fall in temperature, the length contracts, time period decreases, hence clock gains time.<\/p>\n

              \"AP<\/p>\n

              Question 4.
              \nIn what way is the anomalous behaviour of water advantageous to acquatic animals ? [A.P. Mar. 18, Mar. 14]
              \nAnswer:
              \nIn cold countries, as atmospheric temperature decreases, the upper layers of the lakes, rivers etc., cool, contract and sink to the bottom (fig). This goes on until the whole of the water reaches the temperature of 4\u00b0C. When the top layers cool further temperature falls below 4\u00b0C, it expands and becomes lighter. It does not sink downwards and remains at the top. With further cooling the top layer gradually form ice at the top (fig). Ice and water are bad conductors of heat. So the lower layers are protected against freezing by the layers of ice and cold water at 1\u00b0C, 2\u00b0C and 3\u00b0C. This results in water remaining at the bottom at 4\u00b0C. So that aquatic animals survive in those layers of water.
              \n\"AP<\/p>\n

              Question 5.
              \nExplain conduction, convection and radiation with examples. [TS – Mar. ’18, ’16, ’15; AP – Mar. ’15]
              \nAnswer:
              \nThe heat is transmitted in three types. They are 1) Conduction 2) Convection 3) Radiation.<\/p>\n

                \n
              1. Conduction : The process of transmission of heat from one place to other without actual movement of the particles of the medium is called conduction.
                \nEx : When long iron rod is heated at one end, heat transmits to the other end.<\/li>\n
              2. Convection : The process of transmission of heat from one place to another by the actual
                \nmovement of the particles is called convection.
                \nEx. : If water in a beaker is heated, the particles of water at the bottom receive the heat first. These particles expand, become lighter and rise up. At the same time colder and denser particles reach the bottom. They get in their turn heated and move up. This process is known as convection.<\/li>\n
              3. Radiation : The process of transmission of heat from one place to another without any intervening medium is called radiation.
                \nEx. : Earth receives heat radiations from the sun.<\/li>\n<\/ol>\n

                Long Answer Questions<\/span><\/p>\n

                Question 1.
                \nState Boyle’s law and charle’s law. Hence, derive ideal gas equation which of two laws is better for the purpose of Thermometry and why ?
                \nAnswer:
                \nP \u221d \\(\\frac{1}{\\mathrm{V}}\\)
                \n\u21d2 PV = constants \u21d2 P1<\/sub>V1<\/sub> – P2<\/sub>V2<\/sub>
                \nCharles law at constant volume : At constant volume the pressure of given mass of a gas is directly proportional to absolute temperature of gas.
                \nP \u221d T
                \n\u21d2 \\(\\frac{P}{T}\\) = constant \u21d2 \\(\\frac{P_1}{T_1}=\\frac{P_2}{T_2}\\)
                \nCharles law at constant pressure : At constant pressure the volume of given mass of gas is directly proportional to absolute temperature.
                \nV \u221d T
                \n\u21d2 \\(\\frac{V}{T}\\) = constant \u21d2 \\(\\frac{V_1}{T_1}=\\frac{V_2}{T_2}\\)
                \nIdeal gas equation : A gas which obeys all the gas laws at all temperature and all pressures is called an ideal gas.
                \n\"AP
                \nConsider are mole of a gas initially at a pressure P1<\/sub> volume V1<\/sub> and temperature T1<\/sub> and Let P2<\/sub>, V2<\/sub> and T2<\/sub> be the final pressure, volume and absolute temperature T2<\/sub>.
                \nFrom Boyle’s law, P1<\/sub>V1<\/sub> = P2<\/sub>V2<\/sub>
                \n\u21d2 V = \\(\\frac{P_1 V_1}{P_2}\\) ………….. (1)
                \nNow the gas is heated at constant pressure P2<\/sub>, then its volume changes from V to V2<\/sub> and temperature changes from T1<\/sub> and T2<\/sub>.
                \nFrom charles law, \\(\\frac{V}{T_1}=\\frac{V_2}{T_2} \\Rightarrow V=\\frac{V_2 T_1}{T_2}\\) ………….. (2)
                \nFrom (1) & (2) \\(\\frac{P_1 V_1}{P_2}=\\frac{V_2 T_1}{T_2} \\Rightarrow \\frac{P_1 V_1}{T_1}=\\frac{P_2 V_2}{T_2}\\)
                \n\u21d2 \\(\\frac{P V}{T}\\) = constant \u21d2 \\(\\frac{P V}{T}\\) = R
                \nWhere ‘R’ is coniversal gas constant.
                \n\u21d2 PV = RT
                \nFrom ‘n’ gram moles PV = nRT. This law is called ideal gas equation.
                \nOut of Boyle’s law, charles law is better for the purpose of thermometry because with increasing temperature, pressure and volume of gas also increase.<\/p>\n

                \"AP<\/p>\n

                Question 2.
                \nExplain thermal conductivity and coefficient of thermal conductivity. A copper bar of thermal conductivity 401 W\/(mK) has one end at 104\u00b0C and the other end at 24\u00b0C. The length of the bar is 0.10 m and the cross – sectional area is 1.0 \u00d7 10-6<\/sup> m-2<\/sup>. What is the rate of heat conduction, along the bar ?
                \nAnswer:
                \nThermal conductivity : Thermal conductivity of a solid body is ability to conduct heat in the body. Coefficient of thermal conductivity: The coefficient of thermal conductivity of a material may be defined as the quantity of heat energy flowing per second between the opposite faces of the cube of unit side, which are maintained at unit temperature difference.<\/p>\n

                Explanation of conduction : In steady state, conduction between the opposite faces which are at temperatures \u03b81<\/sub>\u00b0C and \u03b82<\/sub>\u00b0C separated by distance d is<\/p>\n

                  \n
                1. Directly proportional to area of cross section of the rod.
                  \nDirectly proportional to temperature difference (\u03b82<\/sub> – \u03b81<\/sub>) between the opposite faces.<\/li>\n
                2. Time of flow of heat, t<\/li>\n
                3. Inversely proportional to the separation of faces ‘d’.
                  \n\"AP
                  \n\u2234 Q \u221d \\(\\frac{\\mathrm{A}\\left(\\theta_2-\\theta_1\\right) t}{\\mathrm{~d}}\\)
                  \nor Q = \\(\\frac{K A\\left(\\theta_2-\\theta_1\\right) t}{d}\\)
                  \nK is called coefficient of thermal conductivity of the material.
                  \nWhen A = 1; (\u03b82<\/sub> – \u03b81<\/sub>) = 1; t = 1; d = 1
                  \n\u2234 K = Q<\/li>\n<\/ol>\n

                  Problem:
                  \nKc<\/sub> = 401 W\/mK, \u03b82<\/sub> = 104\u00b0C, \u03b82<\/sub> = 24\u00b0C, d = 0.10 m, A = 1.6 \u00d7 10-6<\/sup> m-2<\/sup>
                  \nRate of heat conduction = \\(\\frac{Q}{t}=K_c A \\frac{\\left(\\theta_2-\\theta_1\\right)}{d}\\)
                  \n= 401 \u00d7 1 \u00d7 10-6<\/sup> \u00d7 (104-24) = 0.32 W<\/p>\n

                  \"AP<\/p>\n

                  Question 3.
                  \nState and explain Newton’s law of cooling. State the conditions under which Newton’s law of cooling is applicable. A body cools down from 60\u00b0C to 50\u00b0C in 5 minutes and to 40\u00b0C in another 8 minutes. Find the temperature of the surroundings.
                  \nAnswer:
                  \nExpression for Newton’s law of cooling : Consider a hot body at temperature T. Let T0<\/sub> be the temperature of its surroundings. According to Newton’s law of cooling,
                  \nRate of loss of heat \u221d Temperature difference between the body and its surroundings.
                  \n\\(\\frac{-\\mathrm{dQ}}{\\mathrm{dt}}\\) \u221d (T – T0<\/sub>)
                  \n\\(\\frac{-\\mathrm{dQ}}{\\mathrm{dt}}\\) = k (T – T0<\/sub>) \u21d2 (1) where k is proportionality constant. Let m be the mass and C the specific heat of the body at temperature T. If the temperature of the body falls by small amount dT in time dt, then the amount of heat lost is
                  \ndQ = mc dT
                  \n\u2234 Rate of loss of heat is given by
                  \n\\(\\frac{\\mathrm{dQ}}{\\mathrm{dt}}\\) = mc \\(\\frac{\\mathrm{dT}}{\\mathrm{dt}}\\) ………….. (2)
                  \nputting (2) in (1) \u21d2
                  \n-mc \\(\\frac{\\mathrm{dT}}{\\mathrm{dt}}\\) = k(T – T0<\/sub>)
                  \n\\(\\frac{\\mathrm{dT}}{\\mathrm{dt}}=\\frac{\\mathrm{k}}{\\mathrm{mc}}\\) (T – T0<\/sub>) = -K(T – T0<\/sub>)
                  \nwhere K = \\(\\frac{\\mathrm{k}}{\\mathrm{mc}}\\) is another constant.
                  \n\u21d2 \\(\\frac{d T}{\\left(T-T_0\\right)}\\) = – K dt
                  \non integrating both sides, we get
                  \n\\(\\int \\frac{d T}{\\left(T-T_0\\right)}=-K \\int d t\\)
                  \nloge<\/sub> (T – T0<\/sub>) = – Kt + c …………… (3)
                  \nT – T0<\/sub> = e-Kt + c<\/sup>
                  \nT = T0<\/sub> + ec<\/sup> e-Kt<\/sup>
                  \nT = T0<\/sub> + Ce-Kt<\/sup> ……………… (4)
                  \nWhere c is integration constant and C = ec<\/sup>
                  \nequations (1), (2), (3) and (4) are the different expressions for Newton\u2019s law of cooling.<\/p>\n

                  Explanation of Graphs:
                  \n1) If we plot a graph by taking equation (4) different values of temperature difference \u2206T = T – T0<\/sub> along y – axis and the corresponding values of t along X – axis, we get a curve of the form shown in figure. It clearly shows that the rate of cooling is higher initially and then decreases as the temperature of the body falls.
                  \nCurve showing cooling of hot water with time.
                  \n\"AP
                  \n2) More over, the equation (3) is of the form y = mx + c. So if we plot a graph, by taking loge<\/sub>(T – T0<\/sub>) along Y – axis and time t along x – axis, we must get a straight line, as shown in figure. It has a negative slope equal to – K and intercept on Y – axis equal to C.
                  \nIn both of the above situations, Newton’s law of cooling stands verified.
                  \n\"AP
                  \nNewton’s law of cooling is applicable :<\/p>\n

                    \n
                  1. Loss of heat is negligible by conduction and only when it is due to convection.<\/li>\n
                  2. Loss of heat occurs in a stream lined flow of air i.e., forced convection.<\/li>\n
                  3. Temperature of the body is uniformly distributed over it.<\/li>\n
                  4. Temperature differences are moderate i.e., upto 30 K, however if heat body is due to forced convection the law is valid for large differences of temperature also.
                    \n\"AP
                    \n\u2234 Surrounding temperature, \u03b80<\/sub> = \\(\\frac{85^{\\circ} \\mathrm{C}}{3}\\) = 28.33\u00b0C<\/li>\n<\/ol>\n

                    Problems<\/span><\/p>\n

                    Question 1.
                    \nWhat is the temperature for which the readings on Kelvin and Fahrenheit scales are same?
                    \nAnswer:
                    \nThe relation between Kelvin and Fahrenheit scales is given by
                    \n\\(\\frac{K-273.15}{100}=\\frac{F-32}{180}\\)
                    \nbut K = F
                    \n\\(\\frac{F-273.15}{100}=\\frac{F-32}{180}\\)
                    \nF – 273.15 = \\(\\frac{5}{9}\\) F – \\(\\frac{160}{9}\\)
                    \nF – \\(\\frac{5}{9}\\) F = 273.15 – \\(\\frac{160}{9}\\)
                    \n= 273.15 – 17.77
                    \n\\(\\frac{4F}{9}\\) = 255.38
                    \n\u2234 F = \\(\\frac{9}{4}\\) (255.38)
                    \n= 574.6\u00b0F<\/p>\n

                    \"AP<\/p>\n

                    Question 2.
                    \nFind the increase in temperature of aluminium rod if its length \u00a1s to be increased by 1%. (a for aluminium = 25 \u00d7 10-6\/0<\/sup> C). [A.P (Mar. \u201815)]
                    \nAnswer:
                    \nPercentage increase in length = \\(\\frac{l_2-l_1}{l_1}\\) \u00d7 100
                    \n= \u03b1(t2<\/sub> – t1<\/sub>) \u00d7 100 [\u2235 \\(\\frac{l_2-l_1}{l_1}\\) = \u03b1(t2<\/sub> – t1<\/sub>)]
                    \nHere, percentage increase in length = 1,
                    \n\u03b1 = 25 \u00d7 10-6<\/sup>\/ \u00b0C
                    \n1 = 25 \u00d7 10-6<\/sup>(t2<\/sub> – t1<\/sub>) \u00d7 100
                    \nt2<\/sub> – t1<\/sub> = \\(\\frac{1}{25 \\times 10^{-4}}=\\frac{10^4}{25}\\) = 400\u00b0C<\/p>\n

                    Question 3.
                    \nHow much steam at 100\u00b0C is to be passed into water of mass 100 g at 20\u00b0C to raise its temperature by 5\u00b0C ? (Latent heat of steam is 540 cal\/g and specific heat of water is 1 cal\/g\u00b0C).
                    \nAnswer:
                    \nIn the method of mixtures,
                    \nHeat lost by steam = Heat gained by water
                    \nmS<\/sub>LS<\/sub> + mS<\/sub>S(100 – 1) = mW<\/sub>S (t – 20)
                    \nWhere mS<\/sub> is the mass of steam, LS<\/sub> is the latent heat of steam, S is the specific heat of steam and mW<\/sub> is the mass of water.
                    \nHere, LS<\/sub> = 540 cal\/g
                    \nS = 1 cal\/g\u00b0C;
                    \nmW<\/sub> = 100 g
                    \nt = 20 + 5 = 25\u00b0C
                    \nmS<\/sub> \u00d7 540 + mS<\/sub> \u00d7 1 \u00d7 (100 – 25) = 100 \u00d7 1 \u00d7 (25 – 20)
                    \n615 mS<\/sub> = 500
                    \nmS<\/sub> = \\(\\frac{500}{615}\\)
                    \n= 0.813 g.<\/p>\n

                    \"AP<\/p>\n

                    Question 4.
                    \n2 kg of air is heated at constant volume. The temperature of air is increased from 293 K to 313 K. If the specific heat of air at constant volume is 0.718 kJ\/kgK, find the amount of heat absorbed in kJ and kcal.
                    \n(J = 4.2 kJ\/kcal).
                    \nAnswer:
                    \nHere, M = 2 kg
                    \ndT = 313- 293 = 20 K
                    \nCV<\/sub> = 0.718 \u00d7 103<\/sup> J\/Kg – K
                    \nJ = 4.2 KJ\/Kcal
                    \nCV<\/sub> = \\(\\frac{1}{\\mathrm{~m}} \\frac{\\mathrm{d} \\theta}{\\mathrm{dt}}\\)
                    \nd\u03b8 = mCV<\/sub>dT
                    \n= 0.718 \u00d7 103<\/sup> \u00d7 2 \u00d7 20
                    \n= 28720 J
                    \nd\u03b8 = 28.72 KJ
                    \nd\u03b8 = \\(\\frac{28720}{4.2 \\times 10^3}\\)
                    \n= 6.838 Kcal.<\/p>\n

                    Question 5.
                    \nA dock, with a brass pendulum, keeps correct time at 20\u00b0C, but loses 8.212 s per day, when the temperature rises to 30\u00b0C. Calculate the coefficient of linear expansion of brass.
                    \nAnswer:
                    \nHere t1<\/sub> = 20\u00b0C, t2<\/sub> = 30\u00b0C
                    \nLoss of time per day = 8.212 sec
                    \nLoss of time per day = \\(\\frac{1}{2}\\) \u03b1 (t2<\/sub> – t1<\/sub>) \u00d7 86,400
                    \n8.212 = \\(\\frac{1}{2}\\) \u03b1 (30 – 20) \u00d7 86,400
                    \n\u2234 \u03b1 = \\(\\frac{16.424}{864 \\times 10^3}\\)
                    \n= 19 \u00d7 10-6<\/sup>\/C\u00b0<\/p>\n

                    Question 6.
                    \nA body cools from 60\u00b0C to 40\u00b0C in 7 minutes. What will be its temperature after next 7 minutes if the temperature of its surroundings is 10\u00b0C ?
                    \nAnswer:
                    \nRate of cooling \\(\\frac{\\mathrm{d} \\theta}{\\mathrm{dt}}\\) = K(\u03b8av<\/sub> – \u03b80<\/sub>)
                    \nCase (i) :
                    \nGiven, .
                    \n\u03b81<\/sub> = 60\u00b0C, \u03b82<\/sub> = 40\u00b0C, \u03b80<\/sub> = 10\u00b0C, t = 7 min
                    \n\u21d2 \\(\\frac{60^{\\circ} \\mathrm{C}-40^{\\circ} \\mathrm{C}}{7}=K\\left[\\frac{60^{\\circ} \\mathrm{C}+40^{\\circ} \\mathrm{C}}{2}-10^{\\circ} \\mathrm{C}\\right]\\)
                    \n\\(\\frac{20}{7}\\) = K[50 – 10]
                    \n\u21d2 \\(\\frac{20}{7}\\) = K \u00d7 40
                    \n\u2234 K = \\(\\frac{1}{14}\\) ……………. (1)<\/p>\n

                    Case (ii) :
                    \nGiven,
                    \n\u03b81<\/sub> = 40\u00b0C, \u03b80<\/sub> = 10\u00b0C, t = 7 min, \u03b82<\/sub> = ?
                    \n\u21d2 \\(\\frac{40-\\theta_2}{7}=K\\left[\\frac{40+\\theta_2}{2}-10\\right]\\)
                    \n\\(\\frac{40-\\theta_2}{7}=\\frac{1}{14}\\left[\\frac{40+\\theta_2}{2}-10\\right]\\)
                    \n80 – 2\u03b82<\/sub> = \\(\\frac{40+\\theta_2-20}{2}\\)
                    \n160 – 4\u03b82<\/sub> = 20 + \u03b82<\/sub>
                    \n5\u03b82<\/sub> = 140
                    \n\u2234 \u03b82<\/sub> = \\(\\frac{140}{5}\\) = 28\u00b0C<\/p>\n

                    \"AP<\/p>\n

                    Question 7.
                    \nIf the maximum intensity of radiation for a black body is found at 2.65 pm what is the temperature of the radiating body ? (Wien’s constant = 2.9 \u00d7 10-3<\/sup> mK)
                    \nAnswer:
                    \n\u03bbmax<\/sub> = 2.65 pm = 2.65 \u00d7 10-6<\/sup>m
                    \nWien’s constant (b) = 2.90 \u00d7 10-3<\/sup> mk
                    \n\u03bbmax<\/sub> T = b(constant)
                    \nT = \\(\\frac{b}{\\lambda_{\\max }}=\\frac{2.9 \\times 10^{-3}}{2.65 \\times 10^{-6}}\\)
                    \n= 1094 K.<\/p>\n

                    Additional Problems<\/span><\/p>\n

                    Question 1.
                    \nThe triple points of neon and carbon dioxide are 24.57 K and 216.55 K respectively. Express these temperatures on the Celsius and Fahrenheit scales.
                    \nAnswer:
                    \nRelation between kelvin scale and Celsius scale is TC<\/sub> = TK<\/sub> – 273.15
                    \nWhere TC<\/sub>.TK<\/sub> = temperature on Celsius and kelvin scales respectively
                    \nFor Neon TC<\/sub> = 24.57 – 273.15 = – 248.58\u00b0C
                    \nFor CO2<\/sub> TC<\/sub> = 216.55 – 273.15 = -56.60\u00b0C
                    \nRelation between Kelvin and Fahrenheit scales is
                    \n\\(\\frac{T_F-32}{180}=\\frac{T_K-273.15}{100}\\)
                    \nTF<\/sub> = \\(\\frac{180}{100}\\) (TK<\/sub> – 273.15) + 32
                    \nFQ or Neon TF<\/sub>
                    \n= \\(\\frac{180}{100}\\)(24.57 – 273.15) + 32 = – 415.44\u00b0 F
                    \nFQ or CO2<\/sub> TF<\/sub> = \\(\\frac{180}{100}\\) (216.55 – 273.15) + 32
                    \n= – 69.88\u00b0 F<\/p>\n

                    Question 2.
                    \nTwo absolute scales A and B have triple points of water defined to be 200 A and 350 B. What is the relation between TA<\/sub> and TB<\/sub> ?
                    \nAnswer:
                    \nGiven triple point of water on scale A = 200
                    \nA triple point of water on scale B = 350 B
                    \nAs per question 200 A = 350, B = 273.16 K
                    \n(Or)
                    \nIA = \\(\\frac{273.16}{200}\\) K and IB = \\(\\frac{273.16}{350}\\) K
                    \nIf TA<\/sub> and TB<\/sub> represent the triple point of water on two scales A and B then
                    \n\\(\\frac{273.16}{200}\\) TA<\/sub> = \\(\\frac{273.16}{350}\\)TB<\/sub> (Or) \\(\\frac{T_A}{T_B}=\\frac{200}{350}=\\frac{4}{7}\\)
                    \n(Or) TA<\/sub> = \\(\\frac{4}{7}\\) TB<\/sub><\/p>\n

                    \"AP<\/p>\n

                    Question 3.
                    \nThe electrical resistance in ohms of a certain thermometer varies with temperature according to the approximate law:
                    \nR = R0<\/sub>[1 + \u03b1(T – T0<\/sub>)]
                    \nThe resistance is 101.6 \u2126 at the triple-point of water 273.16 K and 165.5 \u2126 at the normal melting point of lead (600.5 K). What is the temperature when the resistance is 123.4 \u2126 ?
                    \nAnswer:
                    \nHere, R0<\/sub> = 101. 6 \u2126, T0<\/sub> = 273.16 K
                    \nCase (i) R1<\/sub> = 165.5 \u2126, T1<\/sub> = 600.5 K
                    \nCase (ii) R2<\/sub> = 123.4 \u2126; T2<\/sub> = ?
                    \nUsing the relation R = R0<\/sub>(1 + \u03b1(T – T0<\/sub>))
                    \nCase (i):
                    \n165.5 = 101.6(1 + \u03b1(600.5-273.16))
                    \n\u03b1 = \\(\\frac{165.5-101.6}{101.6 \\times(600.5-273.16)}\\)
                    \n= \\(\\frac{63.9}{101.6 \\times 327.34}\\)<\/p>\n

                    Case (ii): 123.4 = 101.6(1 + \u03b1(T2<\/sub> – 273.16))
                    \n(Or)
                    \n123.4 = 101.6
                    \n(1 + \\(\\frac{63.9}{101.6 \\times 327.34}\\) (T2<\/sub> – 273.16))
                    \n= 101.6 + \\(\\frac{63.9}{327.34}\\) (T2<\/sub> – 273.16)
                    \nT2<\/sub> = \\(\\frac{(123.4-101.6) \\times 327.34}{63.9}\\) + 273.16
                    \n= 111.67 + 273.16 = 384.83 K<\/p>\n

                    Question 4.
                    \nAnswer the following :
                    \na) The triple-point of water is a standard fixed point in modern thermometry. Why? What is wrong in taking the melting point of ice and the boiling point of water as standard fixed points (as was originally done in the Celsius scale) ?
                    \nb) There were two fixed points in the original Celsius scale as mentioned above which were assigned the number 0\u00b0C and 100\u00b0C respectively. On the absolute scale, one of the fixed points is the triple-point of water, which on the Kelvin absolute scale is assigned the number 273.16 K. What is the other fixed point on this (Kelvin) scale ?
                    \nc) The absolute temperature (Kelvin scale) T is related to the temperature tC<\/sub> on the Celsius scale by
                    \ntc<\/sub> = T – 273.15
                    \nWhy do we have 273.15 in this relation and not 273.16 ?
                    \nAnswer:
                    \na) This is on account of the fact that the triple point of water has a unique value i.e., 273.16 K at a unique point, where exists unique values of pressure and volume. On the other hand, the melting point of ice and boiling point of water do not have unique set of value as they change in pressure and volume.<\/p>\n

                    b) The other fixed point on the absolute scale is the absolute zero itself.<\/p>\n

                    c) On Celsius 0\u00b0C corresponds to melting points of ice of normal pressure. The corresponding value of absolute temperature is 273.15 K. The temperature 273.16 K corresponds to the triple point of water.
                    \nFrom the given relation the corresponding value of triple point of water on Celsius scale
                    \n= 273.16 – 273. 15 = 0.01\u00b0C.<\/p>\n

                    \"AP<\/p>\n

                    Question 5.
                    \nTwo. ideal gas thermometers A and B use oxygen and hydrogen respectively. The following observations are made :
                    \n\"AP
                    \na) What is the absolute temperature of normal melting point of sulphur as read by thermometers A and B?
                    \nb) What do you think is the reason behind the slight difference in answers of thermoineters A and B ? (The thermometers are not faulty). What further procedure is needed in the experiment to reduce the discrepancy between the two readings ?
                    \nAnswer:
                    \na). Let T be the melting point of sulphur, for water Ttr<\/sub> = 273.16 K
                    \nFor thermometer A, T = P\/Ptr<\/sub> \u00d7 273.16
                    \n= \\(\\frac{1.797 \\times 10^5}{1.250 \\times 10^5}\\) \u00d7 273.16 = 392.69 K
                    \nFor thermometer B, T = P\/Ptr<\/sub> \u00d7 273.16
                    \n= \\(\\frac{0.287 \\times 10^5}{0.200 \\times 10^5}\\) \u00d7 273.16 = 391.98 K<\/p>\n

                    b) The cause of slight different answers is that the oxygen and hydrogen gases are not perfectly ideal. To reduce this discrepancy, the readings should be at lower and lower pressure as in that case, the gases approach to the ideal gas behaviour.<\/p>\n

                    Question 6.
                    \nA steel tape 1 m long is correctly calibrated for a temperature of 27.0\u00b0C. The length of a steel rod measured by this tape is found to be 63.0 cm on a hot day when the temperature is 45.0\u00b0C. What is the actual length of the steel rod on that day ? What is the length of the same steel rod on a day when the temperature is 27.0\u00b0C ? Coefficient of linear expansion of steel = 1.20 \u00d7 10-5<\/sup> K-1<\/sup>.
                    \nAnswer:
                    \nLength of steell tape at 27\u00b0C is 100 cm i.e.
                    \nL = 100 cm and T = 27\u00b0C
                    \nThe length of steel tape at 45\u00b0C is L1<\/sup> = L + \u2206L
                    \n= L + \u03b1L\u2206T
                    \n= 100 + (1.20 \u00d7 10-5<\/sup>) \u00d7 100 \u00d7 (45\u00b0 – 27)
                    \n= 100.0216 cm
                    \nLength of 1 cm mark at 27\u00b0C on this scale, at 45\u00b0C
                    \n= 100.0216\/100 cm
                    \nLength of 63 cm measured’ by this tape at 45\u00b0C will be
                    \n= \\(\\frac{100.0216}{100}\\) \u00d7 63 = 63.0136 cm
                    \nLength of the same steel rod on a day when the temperature is 27\u00b0C = 63 \u00d7 1 = 63 cm.<\/p>\n

                    \"AP<\/p>\n

                    Question 7.
                    \nA large steel wheel is to be fitted on to a shaft of the same material. At 27\u00b0C, the outer diameter of the shaft is 8.70 cm and the diameter of the central hole in the wheel is 8.69 cm. The shaft is cooled using ‘dry ice’. At what temperature of the shaft does the wheel slip on the shaft ? Assume coefficient of linear expansion of the steel to be constant over the required temperature range :
                    \n\u03b1steel<\/sub> = 1.20 \u00d7 10-5<\/sup>K-1<\/sup>.
                    \nAnswer:
                    \nHere T1<\/sub> = 27\u00b0C = 27 + 273 = 300 K
                    \nLength at temperature, T1<\/sub>K = LT1<\/sub> = 8.70 cm
                    \nLength at temperature, T2<\/sub>K = LT2<\/sub> – 8.69 cm
                    \nChange in length = LT2<\/sub> – LT1<\/sub> = LT2<\/sub> \u03b1(T2<\/sub> – T2<\/sub>)
                    \nOr 8.69 – 8.70 = 8.70 \u00d7 (1.20 \u00d7 10-5<\/sup>) \u00d7 (T2<\/sub> – 300)
                    \nOr T2<\/sub> – 300 = \\(\\frac{0.01}{8.70 \\times 1.2 \\times 10^{-5}}\\) = -95.8
                    \nOr T2<\/sub> = 300 – 95.8 = 204.2 K = -68.8\u00b0C<\/p>\n

                    Question 8.
                    \nA hole is drilled in a copper sheet. The diameter of the hole is 4.24 cm at 27.0\u00b0C. What is the change in the diameter of the hole when the sheet is heated to 227\u00b0C ? Coefficient of linear expansion of copper = 1.70 \u00d7 10-5<\/sup>K-1<\/sup>.
                    \nAnswer:
                    \nIn this problem superfical expansion of copper sheet will be involved on heating. Here, area of hole at 227\u00b0C, then area of the hole at 227\u00b0C,
                    \nS2<\/sub> = \\(\\frac{\\pi \\mathrm{D}_2^2}{4}\\)cm2<\/sup>.
                    \nCoefficient of superficial expansion of copper is \u03b2 = 2\u03b1 = 2 \u00d7 1.70 \u00d7 10-5<\/sup> = 3.4 \u00d7 10-5<\/sup> c1<\/sup>
                    \nIncrease in area = S2<\/sub> – S1<\/sub> = \u03b2S1<\/sub> \u2206T (or)
                    \nS2<\/sub> = S1<\/sub> + \u03b2S1<\/sub> \u2206T
                    \n= S1<\/sub>(1 + \u03b2\u2206T)
                    \n(Or)
                    \n\\(\\frac{\\pi \\mathrm{D}_2^2}{4}=\\frac{\\pi}{4}\\) (4.24)2<\/sup> [1 + 3.4 \u00d7 10-5<\/sup> (228 – 27)]
                    \n(Or)
                    \nD2<\/sub>2<\/sup> = (4.24)2<\/sup> \u00d7 1.0068
                    \nD2<\/sub> = 4.2544 cm
                    \nChange in diameter = D2<\/sub> – D1<\/sub>
                    \n= 4.2544 – 4.24
                    \n= 0.0144 cm.<\/p>\n

                    Question 9.
                    \nA brass wire 1.8 m long at 27\u00b0C is held taut with little tension between two rigid supports. If the wire is cooled to a temperature of -39\u00b0C, what is the tension developed in the wire, if its diameter is 2.0 mm ? Co-efficient of linear expansion of brass = 2.0 \u00d7 10-5<\/sup>K-1<\/sup>; Young’s modulus of brass = 0.91 \u00d7 1011<\/sup> Pa.
                    \nAnswer:
                    \nHere L = 1.8m, T1<\/sub> = 27\u00b0C, T2<\/sub> = -39\u00b0C, r = 1
                    \nmm = 10-3<\/sup>m, F = ?
                    \n\u03b1 = 2 \u00d7 10-5<\/sup>C-1<\/sup>, y = 0.91 \u00d7 1011<\/sup> N\/m2<\/sup>
                    \nFrom y = \\(\\frac{\\mathrm{FL}}{a \\Delta L}\\), \u21d2 \u2206L = \\(\\frac{\\mathrm{FL}}{\\mathrm{ay}}\\)
                    \nAlso \u2206L = \u03b1L\u2206T = \u2234 \\(\\frac{\\mathrm{FL}}{\\mathrm{ay}}\\) = \u03b1L\u2206T
                    \n(Or) F = \u03b1\u2206Tay = \u03b1(T2<\/sub> – T1<\/sub>) \u03c0r2<\/sup>y
                    \n= 2 \u00d7 10-5<\/sup> \u00d7 (-39 – 27) \u00d7 (10-3<\/sup>)2<\/sup> \u00d7 0.91 \u00d7 1011<\/sup>
                    \n= -3.77 \u00d7 10-2<\/sup> N
                    \nNegative sign indicates that the force is inward due to contraction of the wire.<\/p>\n

                    \"AP<\/p>\n

                    Question 10.
                    \nA brass rod of length 50 cm and diameter 3.0 mm is joined to a steel rod of the same length and diameter. What is the change in length of the combined rod at 250\u00b0C, if the original lengths are at 40.0\u00b0C ? Is there a ‘thermal stress’ developed at the junction ? The ends of the rod are free to expand (Co-efficient of linear expansion of brass = 2,0 \u00d7 10-5<\/sup>K-1<\/sup>, steel = 1.2 \u00d7 10-5<\/sup>K-1<\/sup>.
                    \nAnswer:
                    \n\u2206L1<\/sub> = L1<\/sub>\u03b11<\/sub>\u2206T = 50 \u00d7 (2.10 \u00d7 10-5<\/sup>) (250 – 40) = 0.2205 cm
                    \n\u2206L2<\/sub> = L2<\/sub>\u03b12<\/sub>\u2206T
                    \n= 50 \u00d7 (1.2 \u00d7 10-5<\/sup>) (250 – 40) = 0.216 cm
                    \n\u2234 Change in length of combined rod
                    \n= \u2206L1<\/sub> + \u2206L2<\/sub>
                    \n= 0.220 + 0.126 = 0.346 cm<\/p>\n

                    Question 11.
                    \nThe coefficient of volume expansion of glycerin is 49 \u00d7 10-5<\/sup>K-1<\/sup>. What is the fractional change in its density for a 30\u00b0C rise in temperature ?
                    \nAnswer:
                    \nHere r = 49 \u00d7 10-5<\/sup>C-1<\/sup>, \u2206T = 30\u00b0C .
                    \nAs V1<\/sup> = V + \u2206V = V(1 + r\u2206T)
                    \n\u2234 V1<\/sup> = V(1 + 49 + 10-5<\/sup> \u00d7 30) = 1.0147 V
                    \nAs P = \\(\\frac{m}{V}\\), p1<\/sup> = \\(\\frac{m}{V^1}\\) = \\(\\frac{\\mathrm{m}}{1.0147 \\mathrm{~V}}\\)= 09855 P
                    \nFractional change in density = \\(\\frac{\\rho-\\rho^{\\prime}}{\\rho}\\)
                    \n= \\(\\frac{\\rho-0.9855 \\rho}{\\rho}\\)
                    \n= 0.0145<\/p>\n

                    Question 12.
                    \nA 10 kW drilling machine is used to drill a bore in a small aluminium block of mass 8.0 kg. How much is the rise in temperature of the block in 2.5 minutes, assuming 50% of power is used up in heating the machine itself or lost to the surroundings. Specific heat of aluminium = 0.91 Jg-1<\/sup>K-1<\/sup>.
                    \nAnswer:
                    \nHere p = 10 kw = 105<\/sup>w,
                    \nMass m = 8.0 kg = 8 \u00d7 103<\/sup> g
                    \nRise in temp; \u2206T = ?
                    \ntime t = 2.5 min = 2.5 \u00d7 60 sec
                    \nSp. heat C = 0.91 Jg-1<\/sup>\u00b0C-1<\/sup>
                    \nTotal energy = p \u00d7 t = 104<\/sup> \u00d7 150
                    \n= 15 \u00d7 105<\/sup> J
                    \nAs 50% of energy is lost
                    \n\u2234 Energy available \u2206Q = \\(\\frac{1}{2}\\) \u00d7 15 \u00d7 105<\/sup>
                    \n= 7.5 \u00d7 105<\/sup> J
                    \nAs \u2206Q = mc \u2206T
                    \n\u2234 \u2206T = \\(\\frac{\\Delta Q}{\\mathrm{mc}}=\\frac{7.5 \\times 10^5}{8 \\times 10^3 \\times 0.91}\\) = 103\u00b0C<\/p>\n

                    \"AP<\/p>\n

                    Question 13.
                    \nA copper block of mass 2.5 kg is heated in a furnace to a temperature of 500\u00b0C and then placed on a large ice block. What is the maximum amount of ice that can melt ? (Specific heat of copper = 0.39 Jg-1<\/sup>K-1<\/sup> heat of fusion of water = 335 Jg-1<\/sup>.
                    \nAnswer:
                    \nHere, mass of copper block m = 2.5 kg = 2500 kg
                    \nFall in temperature \u2206T = 500 – 0 = 500\u00b0C
                    \nSpecific heat of copper C = 0.39 Jg-1<\/sup>\u00b0C-1<\/sup>
                    \nLatent heat of fusion L = 335 Jg-1<\/sup>
                    \nLet the mass of ice melted by m1<\/sup>
                    \nAs heat gained by ice = Heat lost by copper
                    \nm-1<\/sup>L = Mc\u2206T
                    \nm-1<\/sup> = \\(\\frac{\\mathrm{mC} \\Delta \\mathrm{T}}{\\mathrm{L}}\\)
                    \nm1<\/sup> = \\(\\frac{2500 \\times 0.39 \\times 500}{335}\\)
                    \n= 1500 g = 1.5 kg<\/p>\n

                    Question 14.
                    \nIn an experiment on the specific heat of a metal, a 0.20 kg block of the metal at 150\u00b0C is dropped in a copper calorimeter (of water equivalent 0.025 kg) containing 150 cm3 of water at 27\u00b0C. The final temperature is 40\u00b0C. Compute the specific heat of the metal. If heat losses to the surroundings are not negligible, is your answer greater or smaller than the actual value for specific heat of the metal ?
                    \nAnswer:
                    \nHere, mass of metal, m = 0.20 kg = 200 g
                    \nFall in temperature of metal \u2206T = 150 – 40
                    \n= 110\u00b0C
                    \nIf L is specific heat of metal, then heat lost by the metal
                    \n\u2206Q = mc\u2206T = 200 \u00d7 L \u00d7 110
                    \nVolume of water = 150 C.C
                    \nMass of water m1<\/sup> = 150 g
                    \nWater equivalent of calorimeter, w = 0.025 kg = 25 kg
                    \nRise in temp of water and calorimeter
                    \n\u2206T1<\/sup> = 40 – 27 = 13\u00b0C
                    \nHeat gained by water and calorimeter,
                    \n\u2206Q1<\/sup> = (m1<\/sup> + w)T1<\/sup>
                    \n= (150 + 25) \u00d7 13 = 175 \u00d7 13
                    \nAs \u2206Q = \u2206Q1<\/sup>
                    \n\u2234 From (i) and (ii)
                    \n200 \u00d7 C \u00d7 110 = 175 \u00d7 13
                    \nC = \\(\\frac{175 \\times 13}{200 \\times 110}\\) \u2248 0.1
                    \n(Or)
                    \nIf some heat is lost to the sorroundings, value of L is so obtained will be less than the actual value of L.<\/p>\n

                    \"AP<\/p>\n

                    Question 15.
                    \nGiven below are observations on molar specific heats at room temperature of some common gases.
                    \n\"AP
                    \nThe measured molar specific heats of these gases are markedly different from those for monatomic gases. Typically, molar specific heat of a monatomic gas is 2.92 cal\/mol K. Explain this difference. What can you infer from the somewhat larger (than the rest) value for chlorine ?
                    \nAnswer:
                    \nThe gases which are listed in the above table are diatomic gases and not mono atomic gases. For diatomic gases, molar specific heat
                    \n= \\(\\frac{5}{2}\\) R = \\(\\frac{5}{2}\\) \u00d7 1.98 = 4.95, which agrees fairly well with all observations listed in the . table except for chlorine. A monoatomic gas molecules has only the transnational motion. A diatomic gas molecules, a part from translation motion. The vibrational as well as rotational motion is also possible. Therefore to raise the temperature of 1 mole of diatomic gas through 1\u00b0C heat is also to be supplied increase not only translation energy but also rotational and vibrational energies. Hence molar specific heat of diatomic gas is greater than that for monoatomic gas. The higher value of molar specific heat of chlorine as compared to hydrogen, nitrogen, oxygen etc. Shows that for chlorine molecule, at room temperature vibrational motion. Also occurs along with translational and rotational motions. Where as other diatomic molecules at room temperature usually have rotational motion apart from their translational motion. This the reason that chlorine has some what larger value of molar specific heat.<\/p>\n

                    Question 16.
                    \nAnswer the following questions based on the P-T phase diagram of carbon dioxide :
                    \na) At what temperature and pressure can the solid, liquid and vapour phases of CO2<\/sub> co-exist in equilibrium ?
                    \nb) What is the effect of decrease of pressure on the fusion and boiling point of CO2<\/sub> ?
                    \nc) What are the critical temperature and pressure for CO2<\/sub> ? What is their significance ?
                    \nd) Is CO2<\/sub> solid, liquid or gas at (a) – 70\u00b0C under 1 atm, (b) -60\u00b0C under 10 atm, (c) 15\u00b0C under 56 atm ?
                    \nAnswer:
                    \na) The solid, liquid and vapour phase of carbon dioxide exist in equilibrium at the point i.e., temperature = -56.6\u00b0C and pressure = 5.11 atm.<\/p>\n

                    b) With the decrease in pressure, both the fusion and boiling point of carbondioxide will decrease.<\/p>\n

                    c) For carbondioxide, the critical tempera-ture is 31,1\u00b0C and critical pressure is 73.0 atm. If the temp of CO2<\/sub> is more than 31.1\u00b0C. It cannot be liquified, however large pressure we may apply.<\/p>\n

                    d) Carbondioxide will be (a) a vapour at – 70\u00b0C under 1 atm (b) a solid at 6\u00b0C under 10 atm (c) a liquid at 15\u00b0C under 56 atm.<\/p>\n

                    \"AP<\/p>\n

                    Question 17.
                    \nAnswer the following questions based on the P – T phase diagram of CO2<\/sub> :
                    \na) CO2<\/sub> at 1 atm pressure and temperature – 60\u00b0C is compressed isothermally. Does it go through a liquid phase ?
                    \nb) What happens when CO2<\/sub> at 4 atm pressure is cooled from room tem-perature at constant pressure ?
                    \nc) Describe qualitatively the changes in a given mass of solid CO2<\/sub> at 10 atm pressure and temperature – 65\u00b0C as it is heated up to room temperature at constant pressure.
                    \nd) CO2<\/sub> is heated to a temperature 70\u00b0C and compressed isothermally. What changes in its properties, do you expect to observe ?
                    \nAnswer:
                    \na) Since the temp – 60\u00b0C lies to the left of 56.6\u00b0C on the curve i.e. lies in the region of vapour and solid phase. So CO2<\/sub> will condense directly into solid with out becoming liquid.<\/p>\n

                    b) Since the pressure 4 atm is less than 5.11 atm. The carbondioxde will condenses directly into solid without becoming liquid.<\/p>\n

                    c) When a solid CO2<\/sub> at 10 atm. Pressure and -65\u00b0C temp is heated, It is first converted into a liquid. A further increase in temperature brings it to the vapour phase. At P = 10 atm. If a horizontal line is drawn parallel to the T-axis. Then the points of intersection of this line with the fusion and vapourisation curve will give the fu-sion and boiling point of CO2<\/sub> at 10 atm.<\/p>\n

                    d) Since 70\u00b0C is higher than the critical temperature of CO2<\/sub>. So the CO2<\/sub> gas can not be converted into liquid state on being compressed isothermally at 70\u00b0C. It will remain in the vapour state. Nowever the gas will depart more and more now its perfect gas behavious with the increase in pressure.<\/p>\n

                    Question 18.
                    \nA child running a temperature of 101 \u00b0F is given an antipyrin (i.e. a medicine that lowers fever) which causes an increase in the rate of evaporation of sweat from his body. If the fever is brought down to 98\u00b0F in 20 min, what is the average rate of extra evaporation caused, by the drug. Assume the evaporation mechanism to be the only way by which heat is lost. The mass of the child is 30 kg. The specific heat of human body is approximately the same as that of water and latent heat of evaporation of water at that temperature is about 580 cal g-1<\/sup>.
                    \nAnswer:
                    \nHere fall in temp = \u2206T = 101 – 98 = 3\u00b0F
                    \n= 3 \u00d7 \\(\\frac{5}{3}\\)\u00b0C = \\(\\frac{5}{3}\\)\u00b0C
                    \nMass of child M = 30 kg
                    \nSp. heat of human body = Sp heat of water
                    \nC = 1000 cal.kg-1<\/sup>C-1<\/sup>
                    \n\u2234 Heat last by the body of child, \u2206Q = mC\u2206T
                    \n= 30 \u00d7 1000 \u00d7 \\(\\frac{5}{3}\\) = 5000 calories
                    \nIf m be the mass of water evapourated in 20 min then m1<\/sup>L = \u2206Q
                    \nm1<\/sup> = \\(\\frac{\\Delta Q}{L}=\\frac{5000}{580}\\) = 86.2 g
                    \n\u2234 Average rate of extra evapouration = \\(\\frac{86.2}{20}\\)
                    \n= 4.31 gmin-1<\/sup><\/p>\n

                    \"AP<\/p>\n

                    Question 19.
                    \nA ‘thermacole’ icebox is a cheap and efficient method for storing small quantities of cooked food in summer in particular. A cubical icebox of side 30 cm has a thickness of 5.0 cm. If 4.0 kg of ice is put in the box, estimate the amount of ice remaining after 6 h. The outside temperature is 45\u00b0C and co-efficient of thermal conductivity of thermacole is 0.01 Js-1<\/sup>m-1<\/sup> K-1<\/sup>. [Heat of fusion of wafer = 335 \u00d7 103<\/sup> J kg-1<\/sup>]
                    \nAnswer:
                    \nHere length of each side l = 30 cm = 0.3 m
                    \nThickness of each side \u2206x = 5 cm = 0.05 m
                    \nTotal surface area through which heat enters into the box
                    \nA = 6 l2<\/sup> = 6 \u00d7 0.3 \u00d7 0.3 = 0 Jum2<\/sup>
                    \nTemp.diff \u2206T = 45 – 0 = 45\u00b0C,
                    \nK = 0.01 JS-1<\/sup>m-1<\/sup>\u00b0C-1<\/sup>
                    \nTime \u2206T = 6 hrs = 6 \u00d7 60 \u00d7 60 S
                    \nLatent heat of fusion L = 335 \u00d7 103<\/sup> J\/kg
                    \nLet m be the mass of ice melted in this time
                    \n\u2206Q = mL = KA \\(\\left(\\frac{\\Delta \\mathrm{T}}{\\Delta \\mathrm{x}}\\right)\\) \u2206t
                    \nm = kA \\(\\left(\\frac{\\Delta T}{\\Delta \\mathrm{T}}\\right) \\frac{\\Delta \\mathrm{t}}{\\mathrm{L}}\\) = 0.01 \u00d7 0.54 \u00d7 \\(\\frac{45}{0.05}\\) \u00d7 \\(\\frac{6 \\times 60 \\times 60}{335 \\times 10^3}\\) = 0.313 kg
                    \nMass of ice left = 4 – 0.313 = 3.687 kg<\/p>\n

                    Question 20.
                    \nA brass boiler has a base area of 0.15 m2<\/sup> and thickness 1.0 cm. It boils water at the rate of 6.0 kg\/min when placed on a gas stove. Estimate the temperature of the part of the flame in contact with the boiler. Thermal conductivity of brass = 109 Js-1<\/sup>m-1<\/sup> K-1<\/sup>: Heat of vaporisation of water = 2256 \u00d7 103<\/sup> Jkg-1<\/sup>.
                    \nAnswer:
                    \nWere A = 0.15 m2<\/sup> \u2206x = 1.0 m = 10-2<\/sup> m
                    \n\\(\\frac{\\Delta Q}{\\Delta t}=\\frac{6 \\times 10^3 \\times 2256}{60}\\) JS-1<\/sup> = 2256 \u00d7 102<\/sup> JS-1<\/sup>
                    \nK = 609 JS-1<\/sup>m-1<\/sup>\u00b0C-1<\/sup>, \u2206T = (t – 100)
                    \nFrom \\(\\frac{\\Delta \\mathrm{Q}}{\\Delta \\mathrm{t}}=\\mathrm{KA}\\left(\\frac{\\Delta \\mathrm{T}}{\\Delta \\mathrm{x}}\\right)\\)
                    \n2256 \u00d7 102<\/sup> = 609 \u00d7 0.15 \\(\\frac{(t-100)}{10^{-2}}\\)
                    \nt – 100 = \\(\\frac{2256}{609 \\times 0.15}\\) = 2470 t
                    \nt = 24.70<\/p>\n

                    \"AP<\/p>\n

                    Question 21.
                    \nExplain why :
                    \na) a body with large reflectivity is a poor emitter.
                    \nb) a brass tumbler feels much colder than a wooden tray on a chilly day.
                    \nc) an optical pyrometer (for measuring high temperatures) calibrated for an ideal black body radiation gives too low a value for the temperature of a red hot iron piece in the open, but gives a correct value for the temperature when the same piece is in the furnace.
                    \nd) the earth without its atmosphere would be inhospitably cold.
                    \ne) healting systems based on circulation of steam are more efficient in warming a building than those based on circulation of hot water.
                    \nAnswer:
                    \na) This is because a body with large refelectivity is a poor absorber of heat and poor absorbers of heat are poor imilters.<\/p>\n

                    b) When we touch a brass tumbler on a chill day; heat flows from our body to the tumbler quickly cas thermal conductivity or brass is very high and as a result, it appears colder. On the other hand as the wood is bad conductor, heat does not flow to the wooden tray from our body, on touching it.<\/p>\n

                    c) When the red hot iron pieces is in the oven, its temperature TK is the given by the relation.
                    \nE = \u03b1T4<\/sup> But if the red hot iron piece is in the open having the surrounding temperature T0<\/sub>, its energy is radiated according to relation E1<\/sup> = \u03b1(T4<\/sup> – T4<\/sup>0<\/sub>). As the working principle of optical pysometer is based on the fact that the brightness of – a glowing surface of a body depends , upon its temperature. Therefore, pysometes gives too low a value for the temperature of red iron in the open.<\/p>\n

                    d) The lower layer’s of Earth’s atmosphere reflect infrared radiations from earth back to the surface of the earth. Thus the heat radiation recieved by the earth from the sun during the day are kept trapped by the atm. If atmosphere of earth were not there, its surface would become too cold to live.<\/p>\n

                    e) Steam at 100\u00b0C possess more heat than the same mass of water at 100\u00b0C possess 540 calories of heat more than possessed by 1 gm of ware at 100\u00b0C. That is why heating systems based on circulation of stream are more efficient than those based on circulation of hot water.<\/p>\n

                    \"AP<\/p>\n

                    Question 22.
                    \nA body cools from 80\u00b0C to 50\u00b0C in 5 minutes. Calculate the time it takes to cool from 60\u00b0C to 30\u00b0C. The temperature of the surroundings is 20\u00b0C.
                    \nAnswer:
                    \n\\(\\frac{\\mathrm{d} \\theta}{\\mathrm{dt}}\\) = K [\u03b8av<\/sub> – \u03b80<\/sub>]
                    \n\\(\\frac{80-50}{5}\\) = K(65 – 20)
                    \n\u21d2 \\(\\frac{30}{5}\\) = K \u00d7 45 = K = \\(\\frac{2}{15}\\)
                    \n\\(\\frac{60-30}{t}=\\frac{2}{15}[45-20]\\)
                    \n\\(\\frac{30}{t}=\\frac{2}{15} \\times 15\\)
                    \nt = 5 min
                    \nTime of cooling is given by t = \\(\\frac{2.3026}{k}\\) log10<\/sub>
                    \n\\(\\frac{T_1-T_0}{T_2-T_0}\\)
                    \nAs per question condition (i) T1<\/sub> = 80\u00b0C, T2<\/sub> = 50\u00b0C, T0<\/sub> = 20\u00b0C, t = 5 min
                    \n5 \u00d7 60 = \\(\\frac{2.3026}{K} \\log _{10} \\frac{80-20}{50-20}\\)
                    \n= \\(\\frac{2.3026}{k} \\log _{10^2}\\) ……………. (2)
                    \nAs per question condition (i) T1<\/sub> = 60\u00b0C, T2<\/sub> = 30\u00b0C, T0<\/sub> = 20\u00b0C, t = ?
                    \nt = \\(\\frac{2.3026}{K} \\log _{10} \\frac{60-20}{30-20}\\)
                    \n= \\(\\frac{2.3026}{k} \\log _{10^4}\\) ……………. (3)
                    \nDiv (3) by (2) we get
                    \n\\(\\frac{t}{5 \\times 60}=\\frac{\\log _{10^4}}{\\log _{10^2}}=\\frac{0.6021}{0.3010}\\) = 2
                    \n(Or) t = 5 \u00d7 60 \u00d7 2 = 10 \u00d7 60 s
                    \n= 10 min<\/p>\n

                    Textual Examples<\/span><\/p>\n

                    Question 1.
                    \nShow that the coefficient of areal expansions. (\u2206A\/A)\/\u2206T, of rectangular sheet of the solid is twice its linear expansivity, \u03b11<\/sub>.
                    \nAnswer:
                    \n\u2206A3<\/sub> = (\u2206a) (\u2206b)
                    \n\"AP
                    \nWhen the temperature increases by \u2206T, a increases by \u2206a = \u03b11<\/sub>a \u2206T and b increases by \u2206b = \u03b11<\/sub>b \u2206T. From fig. the increase in area
                    \n\u2206A = \u2206A1<\/sub> + \u2206A2<\/sub> + \u2206A3<\/sub>
                    \n\u2206A = a \u2206b + b \u2206a + (\u2206a) (\u2206b)
                    \n= a \u03b11<\/sub>b \u2206T + b \u03b11<\/sub>a \u2206T + (\u03b11<\/sub>)2<\/sup> ab(\u2206T)2<\/sup>
                    \n= \u03b11<\/sub>ab \u2206T(2 + \u03b11<\/sub>\u2206T)
                    \n= \u03b11<\/sub>A \u2206T(2 + \u03b11<\/sub>\u2206T)
                    \nSince \u03b11<\/sub> = 10-5<\/sup> K-1<\/sup>, from Table 12.1, the product \u03b11<\/sub>\u2206T for fractional temperature is small in comparision with 2 and may be neglected.
                    \nHence,
                    \n\\(\\left(\\frac{\\Delta \\mathrm{A}}{\\mathrm{A}}\\right) \\frac{1}{\\Delta \\mathrm{T}}\\) \u2248 2\u03b11<\/sub><\/p>\n

                    Question 2.
                    \nA blacksmith fixes iron ring on the rim of the wooden wheel of a bullock cart. The diametbr of the rim and the iron ring are 5.243 m and 5.231 m respectively at 27\u00b0C. To what temperature should the ring be heated so as to fit the rim of the wheel ?
                    \nAnswer:
                    \nGiven,
                    \nT1<\/sub> = 27\u00b0C
                    \nLT1<\/sub> = 5.231 m
                    \nLT2<\/sub> = 5.243 m
                    \nSo,
                    \nLT2<\/sub> = LT1<\/sub> [1 + \u03b11<\/sub>(T2<\/sub> – T1<\/sub>)]
                    \n5.243 m = 5.231 m[1 + 1.20 \u00d7 10-5<\/sup> K-1<\/sup> (T2<\/sub> – 27\u00b0C)]
                    \nor T2<\/sub> = 218\u00b0C<\/p>\n

                    \"AP<\/p>\n

                    Question 3.
                    \nA sphere of aluminium of 0.047 kg is placed for sufficient time in a vessel containing boiling water, so that the sphere is at 100\u00b0C. It is then immediately transfered to 0.14 kg copper calorimeter containing 0.25 kg of water at 20\u00b0C. The temperature of water rises and attains a steady state at 23\u00b0C. Calculate the specific heat capacity of aluminium.
                    \nAnswer:
                    \nMass of aluminium sphere (m.,) = 0.047 kg
                    \nInitial temp, of aluminium sphere = 100\u00b0C
                    \nFinal temp. = 23\u00b0C
                    \nChange in temp. (\u2206T) = (100\u00b0C – 23\u00b0C)
                    \n= 77\u00b0C
                    \nLet specific heat capacity of aluminium be sAl<\/sub>. The amount of heat lost by the aluminium sphere = m1<\/sub>sAl<\/sub> \u2206T = 0.047 kg \u00d7 sAl<\/sub> \u00d7 77\u00b0C s ……………. (i)
                    \nMass of water (m2<\/sub>) = 0.25 kg
                    \nMass of calorimeter (m3<\/sub>) = 0.14 kg
                    \nInitial temp, of water and calorimeter = 20\u00b0C
                    \nFinal temp, of the mixture = 23\u00b0C
                    \nChange in temp. (\u2206T2<\/sub>) = 23\u00b0C – 20\u00b0C = 3\u00b0C
                    \nSpecific heat capacity of water (sw<\/sub>) from table
                    \n12.3 = 4.18 \u00d7 103<\/sup> J kg-1<\/sup> K-1<\/sup>
                    \nSpecific heat capacity of copper calorimeter = 0.386 \u00d7 103<\/sup> J kg-1<\/sup> K-1<\/sup>
                    \nThe amount of heat gained by water and calorimeter = m2<\/sub>sw<\/sub> \u2206T2<\/sub> + m3<\/sub>scu<\/sub> \u2206T2<\/sub> = (m2<\/sub>sw<\/sub> + m3<\/sub>scu<\/sub>) (\u2206T2<\/sub>)
                    \n= (0.25 kg \u00d7 4.18 \u00d7 103<\/sup> J kg-1<\/sup> K-1<\/sup> + 0.14 kg \u00d7 0.386 \u00d7 103<\/sup> J kg-1<\/sup> K-1<\/sup>) (23\u00b0C – 20\u00b0C) ………………… (ii)
                    \nIn the steady state heat lost by the aluminium sphere = heat gained by water + heat gained by calorimeter.
                    \nSo, from (i) and (ii)
                    \n0.047 kg \u00d7 sAl<\/sub> \u00d7 77\u00b0C = (0.25 kg \u00d7 4.18 \u00d7 103<\/sup> J kg-1<\/sup> K-1<\/sup> + 0.14 kg \u00d7 0.386 \u00d7 103<\/sup> J kg-1<\/sup> K-1<\/sup>) (3\u00b0C)
                    \nsAl<\/sub> = 0.911 kJ kg-1<\/sup> K-1<\/sup><\/p>\n

                    Question 4.
                    \nWhen 0.15 kg of ice of 0\u00b0C mixed with 0.30 kg of water at 50\u00b0C in a container, the resulting temperature is 6.7\u00b0C. Calculate the heat of fusion of ice.
                    \n(swater<\/sub> = 4186 J kg-1<\/sup> K-1<\/sup>)
                    \nAnswer:
                    \nHeat lost by water = msw<\/sub> (\u03b8f<\/sub> – \u03b8i<\/sub>)w<\/sub>
                    \n= (0.30 kg) (4186 kg-1<\/sup> K-1<\/sup>) (50.0\u00b0C – 6.7\u00b0C)
                    \n= 54376.14 J
                    \nHeat required to melt ice = m2<\/sub>Lf<\/sub> = (0.15 kg) Lf<\/sub>
                    \nHeat required to raise temperature of ice water to final temperature
                    \n= mI<\/sub>sw<\/sub> (\u03b8f<\/sub> – \u03b8i<\/sub>)I<\/sub>
                    \n= (0.15 kg) (4186 J kg-1<\/sup> K-1<\/sup>) (6.7\u00b0C – 0\u00b0C)
                    \n= 4206.93 J
                    \nHeat lost = heat gained
                    \n54376.14 J = (0.15 Kg)Lf<\/sub> + 4206.93 J
                    \nLf<\/sub> = 3.34 \u00d7 105<\/sup> J kg-1<\/sup><\/p>\n

                    \"AP<\/p>\n

                    Question 5.
                    \nCalculate the heat required to convert 3 kg of ice at -12\u00b0C kept in a calorimeter-to steam at 100\u00b0C at atmospheric pressure. Given specific heat capacity of ice = 2100 J kg-1<\/sup> K-1<\/sup>, specific heat capacity of water = 4186 J kg-1<\/sup> K-1<\/sup>, latent heat of fusion of ice = 3.35 \u00d7 105<\/sup> J kg-1<\/sup> and latent heat of steam = 2.256 \u00d7 106<\/sup> J kg-1<\/sup>.
                    \nAnswer:
                    \nWe have
                    \nMass of the ice, m = 3 kg
                    \nSpecific heat capacity of ice, sice<\/sub>
                    \n= 2100 J kg-1<\/sup> K-1<\/sup>
                    \nSpecific heat capacity of water, swater<\/sub>
                    \n= 4186 J kg-1<\/sup> K-1<\/sup>
                    \nLatent heat of fusion ice, Lf ice<\/sub>
                    \n= 3.35 \u00d7 105<\/sup> J kg-1<\/sup>
                    \nLatent heat of steam, Lsteam<\/sub>
                    \n= 2.256 \u00d7 106<\/sup> J kg-1<\/sup>
                    \nNow,
                    \nQ = Heat required to convert 3 kg of ice at -12\u00b0C to steam at 100\u00b0C.
                    \nQ1<\/sub> = Heat required to convert ice at – 12\u00b0C to ice at 0\u00b0C. .
                    \nmsice<\/sub> \u2206T1<\/sub> = (3 kg) (2100 J kg-1<\/sup> K-1<\/sup>) [0 – (-12)]\u00b0C = 75600 J
                    \nQ2<\/sub> = Heat required to melt ice at – 0\u00b0C to water at 0\u00b0C.
                    \nmLf ice<\/sub> = (3 kg) (3.35 \u00d7 105<\/sup> J kg-1<\/sup>) = 1005000 J
                    \nQ3<\/sub> = Heat required to convert water at 0\u00b0C to water at 100\u00b0C.
                    \nmsW<\/sub> \u2206T2<\/sub> = (3 kg) (4186 J kg-1<\/sup> K-1<\/sup>) (100\u00b0C)
                    \n= 1255800 J
                    \nQ4<\/sub> = Heat required to convert water at 100\u00b0C to steam at 100\u00b0C.
                    \nmLsteam<\/sub> = (3 kg) (2-256 \u00d7 106<\/sup> J kg-1<\/sup>)
                    \n= 6768000 J
                    \nSo,
                    \nQ = Q1<\/sub> + Q2<\/sub> + Q3<\/sub> + Q4<\/sub>
                    \n= 75600 J + 1005000 J + 1255800 J + 6768000 J
                    \n= 9.1 \u00d7 106<\/sup> J<\/p>\n

                    Question 6.
                    \nWhat is the temperature of the steel- copper junction in the steady state of the system shown in fig. Length of the steel rod = 15.0 cm, length of the copper rod = 10.0 cm, temperature of the furnace = 300\u00b0C, temperature of the other end = 0\u00b0C. The area of cross section of the steel rod is twice that of the copper rod. (Thermal conductivity of steel = 50.2 J s-1<\/sup>m-1<\/sup>K-1<\/sup> and of copper = 385 J s-1<\/sup>m-1<\/sup>K-1<\/sup>].
                    \n\"AP
                    \nAnswer:
                    \nLet T be the temperature of the steel-copper junction in the steady state. Then,
                    \n\\(\\frac{K_1 A_1(300-T)}{L_2}=\\frac{K_2 A_2(T-0)}{L_2}\\)
                    \nWhere 1 and 2 refer to the steel and copper rod respectively. For A1<\/sub> = 2, A2<\/sub>, L1<\/sub> = 15.0 cm, L2<\/sub> = 10.0 cm, K1<\/sub> = 50.2 J s-1<\/sup>m-1<\/sup>K-1<\/sup>,
                    \nK2<\/sub> = 385 J s-1<\/sup>m-1<\/sup>K-1<\/sup>, we have
                    \n\\(\\frac{50.2 \\times 2(300-\\mathrm{T})}{15}=\\frac{385 \\mathrm{~T}}{10}\\)
                    \nwhich gives T = 44.4\u00b0C<\/p>\n

                    \"AP<\/p>\n

                    Question 7.
                    \nAn iron bar (L1<\/sub> = 0.1 m, A1<\/sub> = 0.02 m2<\/sup>, K1<\/sub> = 79 W m-1<\/sup> K-1<\/sup>) and a brass bar (L2<\/sub> = 0.1 m, A2<\/sub> = 0.02 m2<\/sup>, K2<\/sub> = 109 W m-1<\/sup>K-1<\/sup>) are soldered end to end as shown in fig. The free ends of the iron bar and brass bar are maintained at 373 K and 273 K respectively. Obtain expressions for and hence compute (i) the temperature of the junction of the two bars, (ii) the equivalent thermal conductivity of the compound bar and (iii) the heat current through the compound bar.
                    \n\"AP
                    \nAnswer:
                    \nGiven,
                    \nL1<\/sub> = L2<\/sub> = L = 0.1 m, A1<\/sub> = A2<\/sub> = A = 0.02 m2<\/sup>, K1<\/sub> = 79 W m-1<\/sup> K-1<\/sup>, K2<\/sub> = 109 W m-1<\/sup>K-1<\/sup>, T1<\/sub> = 373 K and T2<\/sub> = 273 K.
                    \nUnder steady state condition, the heat current (H1<\/sub>) through iron bar is equal to the heat current (H2<\/sub>) through brass bar.
                    \nSo, H = H1<\/sub> = H2<\/sub>
                    \n= \\(\\frac{K_1 A_1\\left(T_1-T_0\\right)}{L_1}=\\frac{K_2 A_2\\left(T_0-T_2\\right)}{L_2}\\)
                    \nFor A1<\/sub> = A2<\/sub> = A and L1<\/sub> = L2<\/sub> = L this equation leads to
                    \nK1<\/sub>(T1<\/sub> – T0<\/sub>) = K2<\/sub>(T0<\/sub> – T2<\/sub>)
                    \nThus the junction temperature T0<\/sub> of the two bars is T0<\/sub> = \\(\\frac{\\left(K_1 T_1+K_2 T_2\\right)}{\\left(K_1+K_2\\right)}\\) ………….. (a)
                    \nUsing this equation, the heat current H through either bar is
                    \n\"AP
                    \nUsing these equations, the heat current H\u2019 through the compound bar of length L1<\/sub> + L2<\/sub> = 2L and the equivalent thermal conductivity K, of the compound bar are given by
                    \n\"AP
                    \n\"AP<\/p>\n

                    \"AP<\/p>\n

                    Question 8.
                    \nA pan filled with hot food cools from 94\u00b0C to 86\u00b0C in 2 minutes when the room temperature is at 20\u00b0C. How long will it take to cool from 71\u00b0C to 69\u00b0C ?
                    \nAnswer:
                    \nThe average temperature of 94\u00b0C and 86\u00b0C is 90\u00b0C, which is 70\u00b0C above the room temperature. Under these conditions the pan cools 8\u00b0C in 2 minutes.
                    \nUsing \\(\\frac{\\mathrm{dT}_2}{\\mathrm{~T}_2-\\mathrm{T}_1}=-\\frac{\\mathrm{k}}{\\mathrm{ms}}\\) dt = – K dt we have
                    \nThe average of 69\u00b0C and 71\u00b0C is 70\u00b0C, which is 50\u00b0C above room temperature. K is the same for this situation as for the original.
                    \n\"AP
                    \nThe average of 69\u00b0C and 71\u00b0C is 70\u00b0C, which is 50\u00b0C above room temperature. K is the same for this situation as for the original.
                    \n\\(\\frac{2^{\\circ} \\mathrm{C}}{\\text { Time }}\\) = K(50\u00b0C)
                    \nWhen we divide above two equations, we have
                    \n\\(\\frac{8^{\\circ} \\mathrm{C} \/ 2 \\min }{2^{\\circ} \\mathrm{C} \/ \\text { time }}=\\frac{K\\left(70^{\\circ} \\mathrm{C}\\right)}{\\mathrm{K}\\left(50^{\\circ} \\mathrm{C}\\right)}\\)
                    \nTime = 0.7 min = 42 s<\/p>\n","protected":false},"excerpt":{"rendered":"

                    Andhra Pradesh BIEAP\u00a0AP Inter 1st Year Physics Study Material 12th Lesson Thermal Properties of Matter Textbook Questions and Answers. AP Inter 1st Year Physics Study Material 12th Lesson Thermal Properties of Matter Very Short Answer Questions Question 1. Distinguish between heat and temperature. [T.S. Mar. 15] Answer: Heat It is a form of energy. It …<\/p>\n","protected":false},"author":5,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"spay_email":""},"categories":[15],"tags":[],"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/posts\/35814"}],"collection":[{"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/users\/5"}],"replies":[{"embeddable":true,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/comments?post=35814"}],"version-history":[{"count":1,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/posts\/35814\/revisions"}],"predecessor-version":[{"id":35815,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/posts\/35814\/revisions\/35815"}],"wp:attachment":[{"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/media?parent=35814"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/categories?post=35814"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/tags?post=35814"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}