 Heat Transfer

Concept Of Physics

H C Verma

1   A uniform slab of dimension $10 cm \times 10 cm \times 1 cm$ is kept between two heat reservoirs at temperatures $10^o$C and $90^o$C. The larger surface areas touch the reservoirs. The thermal conductivity of the material is 0'80 W/m-°C. Find the amount of heat flowing through the slab per minute.

Solution : $t_1 = 90^{\circ}C, \qquad t_2 = 10^{\circ}C$ $\\$ $l = 1cm = 1 \times 10^{-3} m.$ $\\$ $A = 10cm \times 10cm = 0.1 \times 0.1 m^2 = 1 \times 10^{-2} m^2$ $\\$ $K = 0.80 w/m-^{\circ}C$ $\\$ $\frac{Q}{t} = \frac{KA(\theta_1 - \theta_2)}{l} = \frac{8 \times 10^{-1} \times 1 \times 10^{-2} \times 80}{1 \times 10^{-2}} = 64 J/s = 64 \times 60 = 3840J$

2   A liquid-nitrogen container is made of a 1 cm thick thermocoal sheet having thermal conductivity $0'025 J/m-s-^{\circ}C$. Liquid nitrogen at 80 K is kept in it. A total area of 0 - S0 m 3 is in contact with the liquid nitrogen. The atmospheric temperature is 300 K. Calculate the rate of heat flow from the atmosphere to the liquid nitrogen.

Solution :

$t= 1cm = 0.01m, \qquad A = 0.8 m^2$ $\\$ $\theta_1 = 300$ $\\$ $K = 0.025,$ $\\$ $\frac{Q}{t} = \frac{KA(\theta_1 - \theta_2)}{l} = \frac{0.025 \times 0.8 \times (30030)}{0.01} = 440$ watt $\\$

3   The norma! body-temperature of a person is $97^{\circ}$F. Calculate the rate at which heat is flowing out of his body through the clothes assuming the following values. Room temperature -$47^{\circ}F$, surface of the body under clothes = 1 '6 m \ conductivity of the cloth = $0'04 J/m—s—^{\circ}C$. thickness of the cloth = 0 5 cm.

Solution :

$K = 0.04J/m - 5^{\circ}C, \qquad A = 1.6 m^2$ $\\$ $t_1 = 97^{\circ}F = 36.1^{\circ}C \qquad t_2 = 47^{\circ}F = 8.33^{\circ}C$ $\\$ $\frac{Q}{t} = \frac{KA(\theta_1 - \theta_2)}{l} = \frac{4 \times 106{-2} \times 1.6 \times 27.78}{5 \times 10^{-3} } = 356 J/s$

4   Water is boiled in a container having a bottom of surface area 25 cm \ thickness l'O mm and thermal conductivity $50 W/m-^oC$. 100 £ of water is converted into steam per minute in the steady state after the boiling starts. Assuming that no heat is lost to the atmosphere, calculate the temperature of the lower surface of the bottom. Latent heat of vaporization of water = $2'26 x 10$J/ kj:

Solution :

$A = 25 cm^2 = 25 \times 10^{-4} m^2$ $l = 1 mm = 10^{-3} m$ $\\$ $K = 50 w/m-^{\circ}C$ $\\$ $\frac{Q}{t} =$Rate of conservsstion of water into steam $\\$ $=\frac{100 \times 10^{-3} \times 2.26 \times 10^{6}}{1 min} = \frac{10^{-1} \times 2.26 \times 10^6}{60} = 0.376 \times 10^4$ $\\$ $\frac{Q}{t} = \frac{KA(\theta_1 - \theta_2)}{l} = 0.376 \times 10^4 = \frac{50 \times 25 \times 10^{-4} \times (\theta - 100)}{10^{-3}}$ $\\$ $\qquad \Rightarrow \theta = \frac{10^{-3} \times 0.376 \times 10^4}{50 \times \times 25 \times 10^{-4}} = \frac{10^5 \times 0.376}{50 \times 25} = 30.1 = 30$

5   One end of a >:eel rod $(A - 46 J'm-s-^°C)$ of length l'O m is kept in ice it CV ar.i the other end is kept in boiling water at $10J^°$C. The area of cross-section of the rod is 0'04 err. ''. Assuming: so heat loss to the atmosphere, find the mass of the i-v melting per second. Latent heat of fusion of ice - S'or x 10 ' J/kg.

Solution : $K = 46 w/m -s^{\circ}C$ $\\$ $l = 1 m$ $\\$ $A = 0.04 cm^2 = 4 \times 10^{-6} m^2$ $\\$ $L_{fussion ice} = 3.36 \times 10^5 j/kg$ $\\$ $\frac{Q}{t} = \frac{46 \times 4 \times 100 \times 10^{-6}}{1} = 5.4 \times 10^{-8}kg = 5.4 \times 10^{-5}g$

6   An icebox almost completely filled with ice at $0^°$C is dipped into 'arev volume of water at $20^°$C. The box has walls of surface area 2400 cm 2 , thickness 2*0 mm and thermal conductivity $0*06 W/m-^°$C. Calculate the rate at which the ice melts in the box. Latent heat of fusion of ice = 3*4 * 10 J/kg.

Solution :

$A = 2400 cm^2 = 2400 \times 10^{-4}m^2$ $\\$ $l = 2mm = 2\times 10^{-3}m$ $\\$ $K =0.06 w/m-^{\circ}C$ $\\$ $\theta_1 = 20^{\circ}C$ $\\$ $\theta_2 = 0^{\circ}C$ $\\$ $\frac{Q}{t} = \frac{KA(\theta_1 - \theta_2)}{l} = \frac{0.06 \times 2400 \times 10^{-4} \times 20}{2 \times 10^{-3}} = 24 \times 6 \times 10^{-1} \times 10 = 24 \times 6 = 144 J/sec$ $\\$ Rate in which ice melts = \frac{m}{t} = \frac{Q}{t \times L} = \frac{144}{3.4 \times 10^5}Kg/h = \frac{144 \times 3600}{3.4 \times 10^{5}} = 1.52 kg/s 

7   A pitcher with 1 mm thick porous walls contains 10 kg of water. Water comes to its outer surface and evaporates at the rate of O'l g/s. The surface area of the pitcher (one side) «* 200 cm 2 . The room temperature = $42^°$C, latent heat of vaporization - 2*27 * 10 0 J/kg, and the thermal conductivity of the porous walls =$0 80 J/m-s-^°$C. Calculate the temperature of water in the pitcher when it attains a constant value.

Solution :

$l = 1 mm = 10^{-3}m, \qquad m =10kg$ $\\$ $A = 200cm^2 = 2 \times 10^{-2} m^2$ $\\$ $L_{vap} = 2.27 \times 10^{6}J/Kg$ $\\$ $K= 0.80 J/m-s-^{\circ}C$ $\\$ $dQ = 2.27 \times 10^{6} \times 10$ $\\$ $\frac{dQ}{dt} = \frac{2.27 \times 10^7}{10^5} = 2.27 \times 10^2 J/s$ $\\$ Again we know, $\\$ $\frac{dQ}{t} =\frac{0.80 \times 2 \times 10^{-2} \times (42 - T)}{1 \times 10^{-3}}$ $\\$ So, $\frac{8 \times 2 \times 10^{-3} (42 - T)}{10^{-3}} = 2.27 \times 10^{2}$ $\\$ $\Rightarrow 16 \times 42 -16T = 227 \quad \Rightarrow T = 27.8 = 28^{\circ}C$

8   A steel frame (K = $45 W/m-^°$C) of total length 60 cm and - cross-sectional area 0*20 $cm^2$ , forms three sides of a square. The free ends are maintained at $20^°$C and $40^°$C. Find the rate of heat flow through a cross-section of the frame.

Solution : $\\$ $K = 45 w/m-^{\circ}C$ $\\$ $l = 60 cm =60 \times 10^{-2}m$ $\\$ $A = 0.2 cm^2 = 0.2 \times 10^{-4} m^2$ $\\$ $Rate of heat flow,$ $\\$ $= \frac{KA(\theta_1 -\theta_2)}{l} = \frac{45 \times 0.2 \times 10^{-4} \times 20}{60 \times 10^{-2}} = 30 \times 10^{-3} = 0.03 W$

9   Water at $50^°$C is filled in a closed cylindrical vessel of height 10 cm and cross-sectional area 10 $cm ^2$ . The walls of the vessel are adiabatic but the flat parts are made of 1 mm thick aluminium (K «$200 J/m-s-^°$C). Assume that the outside temperature is $20^°$C. The density of water is 1000 kg/m and the specific heat capacity of water = $4200 JAg-^°$C. Estimate the time taken for the temperature to fall by $1 - 0^°$C. Make any simplifying assumptions you need but specify them.

Solution :

$\\$ $A = 10cm^2, \qquad h = 10cm$ $\\$ $\frac{\Delta Q}{\Delta t} = \frac{KA(\theta_1 -\theta_2)}{l} = \frac{200 \times 10^{-3} \times 30 }{1 \times 10^{-3}} = 6000$ $\\$ Since, heat goes out from both surfaces. Hence net heat coming out. $\\$ $\frac{\Delta Q}{\Delta t} = 6000 \times 2 = 12000, \qquad \frac{\Delta Q}{\Delta t} = MS\frac{\Delta \theta}{\Delta t}$ $\\$ $\Rightarrow 6000 \times 2 = 10^{-3} \times 10^{-1} \times 1000 \times 4200 \times \frac{\Delta \theta}{\Delta t}$ $\\$ $\Rightarrow \frac{\Delta \theta}{\Delta t} = \frac{72000}{420} = 28.57$ $\\$ So, in 1 sec $28.57^{\circ}C$ is dropped. $\\$ Hence for drop of $1^{\circ}C \frac{1}{28.57} sec = 0.035 sec$ is required.

10   The left end of a copper rod (length - 20 cm, area of cross-section = 0'20 cm 2 ) is maintained at $20^°$C and the right end is maintained at $80^°$C. Neglecting any loss of heat through radiation, find (a) the temperature at a point 11 cm from the left end and (b) the heat current through the rod. Thermal conductivity of copper = $385 W/m-^°$C.

Solution : $l = 20cm = 20 \times 10^{-2}m$ $\\$ $A = 0.2cm^2 = 0.2 \times 10^{-4}m^2$ $\\$ $\theta_1 = 80^{\circ}C, \qquad \theta_2 = 20^{\circ}C, \qquad K = 385$ $\\$ (a) $\frac{Q}{t} = \frac{KA(\theta_1 -\theta_2)}{l} = \frac{385 \times 0.2 \times 10^{-4}(80-20)}{20 \times 10^{-2}} = 385 \times 6 \times 10^{-4} \times 10 = 2310 \times 10^{-3} = 2.31$ $\\$ (b) Let the temp of the 11 cm point be $\theta$ $\\$ $\frac{\Delta \theta}{\Delta l} = \frac{Q}{tKA}$ $\\$ $\Rightarrow \frac{\Delta \theta}{\Delta l} = \frac{2.31}{385 \times 0.2 \times 10^{-4}}$ $\\$ $\Rightarrow \frac{\theta - 20}{11 \times 10^{-2}} = \frac{2.31}{385 \times 0. \times 10^{-4}}$ $\\$ $\Rightarrow \theta -20 = \frac{2.31 \times 10^4}{385 \times 0.2} \times 11 \times 10^{-2} = 33$ $\\$ $\Rightarrow \theta = 33 + 20 = 53.$

11   The ends of a metre stick are maintained at $100^°$C and $0^°$C. One end of a rod is maintained at $25^°$C. Where should its other end be touched on the metre stick so that there is no heat current in the rod in steady state ?

Solution : Let teh point to be touched be 'B' $\\$ No heat will flow when, the temp at that point is also $25^{\circ}C \quad i.e. \quad Q_{AB} = Q_{BC}$ $\\$ So, $\frac{KA(100-25)}{100- x} = \frac{KA(25-0)}{x}$ $\\$ $\Rightarrow 75x = 2500 - 25x \quad \Rightarrow 2500 \quad \Rightarrow x = 25 cm$ from th end with $0^{\circ}C$

12   A cubical box of volume $216 cm ^3$ is made up of OT cm thick wood. The inside is heated electrically by a 100 W heater. It is found that the temperature difference between the inside and the outside surface is $5^°$C in steady state. Assuming that the entire electrical energy spent appears as heat, find the thermal conductivity of the material of the box.

Solution : $\\$ $V = 216 cm^3$ $\\$ $a = 6cm, \qquad Surface area = 6a^2 = 6 \times 36 m^2$ $\\$ $t = 0.1 cm \qquad \frac{Q}{t} = 100W,$ $\\$ $\frac{Q}{t} = \frac{KA(\theta_1 -\theta_2)}{l}$ $\\$ $\Rightarrow 100 = \frac{K \times 6 \times 36 \times 10^{-4} \times 5}{0.1 \times 10^{-2}}$ $\\$ $\Rightarrow K = \frac{100}{6 \times 36 \times 5 \times 10^{-1}} = 0.9259 W/m^{\circ}C = 0.92W/m^{\circ}C$

13   Figure ($28-E1)$ shows water in a container having 2'0 mm thick walls made of a material of thermal conductivity $0*50 W/m-^°$C. The container is kept in a melting-ice bath at $0^°$C. The total surface area in contact with water is $0'05 m ^2$ . A wheel is clamped inside the water and is coupled to a block of mass M as shown in the figure. As the block goes down, the wheel rotates. It is found that after some time a steady state is reached in which the block goes down with a constant speed of 10 cm/s and the temperature of the water remains constant at $1'0^°C$. Find the mass M of the block. Assume that the heat flows out of the water only through the walls in contact. Take g - $10 m/s ^2$

Solution : $Given \quad \theta_1 = 1^{\circ}C, \qquad \theta_2 =0^{\circ}C$ $\\$ $K = 0.50 w/m-^{\circ}C, \qquad d=2mm = 2 \times 10^{-3}m$ $\\$ $A =5 \times 10^{-2}m^2, \qquad v = 10cm/s = 0.1m/s$ $\\$ $Power = Force \times Velocity = Mg \times v$ $\\$ Again Power $= \\frac{dQ}{dt} = \frac{KA(\theta_1 -\theta_2)}{d}$ $\\$ So, $Mgv = \frac{KA(\theta_1 -\theta_2)}{d}$ $\\$ $\Rightarrow M = \frac{KA(\theta_1 -\theta_2)}{dvg} = \frac{5 \times 10^{-1} \times 5 \times 10^{-2} \times 1}{ 2 \times 10^{-3} \times 10^{-1} \times 10} = 12.5 kg$

14   On a winter day when the atmospheric temperature drops to -$10^°$C, ice forms on the surface of a lake, (a) Calculate the rate of increase of thickness of the ice when 10 cm of ice is already formed, (b) Calculate the total time taken in forming 10 cm of ice. Assume that the temperature of the entire water reaches $0^°$C before the ice starts forming. Density of water = 1000 kg/m latent heat of fusion of ice = 3-36 x 10 J/kg and thermal conductivity of ice = $1*7 W/m-^°$C. Neglect the expansion of water on freezing.

Solution : $K = 1.7 W/m-^{\circ}C, \qquad f_w = 1000 Kg/m^3$ $\\$ $L_{ice} = 3.36 \times 10^5 J/Kg, \qquad T = 10cm = 10 \times 10^{-2}m$ $\\$ (a)$\frac{Q}{t} = \frac{KA(\theta_1 -\theta_2)}{l} \Rightarrow \frac{l}{t} = \frac{KA(\theta_1 -\theta_2)}{Q} = \frac{KA(\theta_1 -\theta_2)}{mL}$ $\\$ $= \frac{KA(\theta_1 -\theta_2)}{AtF_wL} = \frac{1.7 \times [(0 - (-10))]}{10 \times 10^{-2} \times 1000 \times 3.36 \times 10^{5}}$ $\\$ $= \frac{17}{3.36} \times 10^{-7} = 5.059 \times 10^{-7} = 5 \times 10^{-7} m/sec$ $\\$ (b) let us assume that c length of ice has become formed to form a small strip of ice if length dx, dt time is required. $\\$ $\frac{dQ}{dt} = \frac{KA(\Delta \theta)}{x} \Rightarrow \frac{dmL}{dt} = \frac{KA(\Delta \theta)}{x} \Rightarrow \frac{AdxfwL}{dt} = \frac{KA(\Delta \theta)}{x}$ $\\$ $\Rightarrow \frac{dxfwL}{dt} = \frac{KA(\Delta \theta)}{x} \Rightarrow dt =\frac{KA(xdxfwL)}{K(\Delta \theta)}$ $\\$ $\Rightarrow \int_{0}^{1}dt = \frac{fwL}{K(\Delta \theta)}\int_{0}^{1}xdx \qquad \Rightarrow t = \frac{fwL}{K(\Delta \theta)}\Bigg[ \frac{x^2}{2} \Bigg]_{0}^{1} = \frac{fwL}{K(\Delta \theta)}\frac{l^2}{2}$ $\\$Putting the values, $\\$ $\Rightarrow t= \frac{1000 \times 3.36 \times 10^5 \times (10 \times 10^-2)^2}{1.7 \times 10 \times 2} = \frac{3.36 }{2 \times 17} \times 10^6sec = \frac{3.36 \times 10^6}{2 \times 17 \times 3600}hrs = 27.45 hrs = 27.5 hrs.$

15   Consider the situation of the previous problem. Assume that the temperature of the water at the bottom of the lake remains constant at $4^°$C as the ice forms on the surface (the heat required to maintain the temperature of the bottom layer may come from the bed of the lake). The depth of the lake is l'O m. Show that the thickness of the ice formed attains a steady state maximum value. Find this value. The thermal conductivity of water = $0'50 W/m—^°$C. Take other relevant data from the previous problem.

Solution : Let the 'B' be the maximum level upto which ice is formed. Hence the heat conducted at that point from both the levels is the same.$\\$ Let $AB = x$ $\\$ $i.e. \frac{Q}{t}ice = \frac{Q}{t}water \qquad \frac{K_ice \times A \times 10}{x} = \frac{K_water \times A \times 4}{(1-x)}$ $\\$ $\Rightarrow \frac{1.7 \times 10}{x} = \frac{5 \times 10^{-1} \times 4}{1-x} \Rightarrow \frac{17}{x} = \frac{2}{1-x}$ $\\$ $17 - 17x = 2x \Rightarrow 19x = 17 \Rightarrow x = \frac{17}{19} = 0.894 = 89 cm$

16   Three rods of lengths 20 cm each and area of cross-section 1 cm " are joined to form a triangle ABC. The conductivities of the rods are K $Att = 50 J/m-s-°$C, K ll( .$= 200 J/m-s-°$C and K $AC = 400 J/m-s-^°$C. The junctions A, B and C are maintained at $40^°$C, 80°C and $80^°$C respectively. Find the rate of heat flowing through the rods AB, AC and BC.

Solution : $K_{AB} = 50 j/m-s^{\circ}C \qquad \theta_A = 40^{\circ}C$ $\\$ $K_{BC} = 200 j/m-s^{\circ}C \qquad \theta_B = 80^{\circ}C$ $\\$ $K_{AC} = 400 j/m-s^{\circ}C \qquad \theta_C = 80^{\circ}C$ $\\$ Length = $20 cm = 20 \times 10^{-2}m$ $\\$ $A = 1cm^2 = 1 \times 10^{-4}m^2$ $\\$ $(a) \frac{Q_{AB}}{t} = \frac{K_{AB} \times A(\theta_B - \theta_A)}{l} = \frac{50 \times 1 \times 10^{-4} \times 40}{20 \times 10^{-2}} = 1W$ $\\$ $(b) \frac{Q_{AC}}{t} = \frac{K_{AC} \times A(\theta_C - \theta_A)}{l} = \frac{400 \times 1 \times 10^{-4} \times 40}{20 \times 10^{-2}} = 800 \times 10^{-2} = 8$ $\\$ $(c) \frac{Q_{BC}}{t} = \frac{K_{BC} \times A(\theta_B - \theta_C)}{l} = \frac{200 \times 1 \times 10^{-4} \times 0}{20 \times 10^{-2}} = 0$

17   A semicircular rod is joined at its end to a straight rod of the same material and the same cross-sectional area. The straight rod forms a diameter of the other rod. The junctions are maintained at different temperatures. Find the ratio of the heat transferred through a cross-section of the semicircular rod to the heat transferred through a cross-section of the straight rod in a given time.

Solution : We know $Q = \frac{KA(\theta_1 - \theta_2)}{d}$ $\\$ $Q_1 = \frac{KA(\theta_1 - \theta_2)}{d_1}, \qquad Q_2 = \frac{KA(\theta_1 - \theta_2)}{d_2}$ $\\$ $\frac{Q_1}{Q_2} = \frac{\frac{KA(\theta_1 - \theta_2)}{\pi r}}{\frac{KA(\theta_1 - \theta_2)}{2r}} = \frac{2r}{\pi r} = \frac{2}{\pi} \qquad [d_1 = \pi r, \quad d_2 = 2r]$

18   A metal rod of cross-sectional area 1*0 cm 2 is being heated at one end. At one time, the temperature gradient is $5*0^°$C/cm at cross-section A and is $2*5^°$C/cm at cross-section B. Calculate the rate at which the temperature is increasing in the part AB of the rod. The heat capacity of the part AB - $0*40 J/^°$C, thermal conductivity of the material of the rod - $200 W/m-^°$C. Neglect any loss of heat to the atmosphere.

Solution :

The rate of heat floew per sec $= \frac{dQ_A}{dt} = KA \frac{d \theta}{dt}$ $\\$ The rate of heat floew per sec $= \frac{dQ_B}{dt} = KA \frac{d \theta_B}{dt}$ $\\$ This part of heat is absorbed by the red. $\\$ $\frac{Q}{t} = \frac{ms\Delta \theta}{dt}, \qquad \frac{d \theta}{dt} = Rate of net temp. variation$ $\\$ $\Rightarrow \frac{msd \theta}{dt} = KA\frac{d\theta_A}{dt} - KA\frac{d\theta_B}{dt} \Rightarrow ms \frac{d\theta}{dt} = KA\Bigg( \frac{d\theta_A}{dt} -\frac{d\theta_B}{dt}\Bigg)$ $\\$ $\Rightarrow 0.4 \times \frac{d\theta}{dt} = 200 \times 1 \times 10^{-4} (5 - 2.5)^{\circ}C/cm$ $\\$ $\Rightarrow 0.4 \times \frac{d\theta}{dt} = 200 \times 2.5 \times 10^{-4}$ $\\$ $\Rightarrow \frac{d\theta}{dt} = \frac{200 \times 2.5 \times 10^{-4}}{0.4 \times 10^{-2}}^{\circ}C/m = 1250 \times 10^{-2} = 12.5^{\circ}C/m$

19   Steam at $120^°$C is continuously passed through a 50 cm long rubber tube of inner and outer radii l'C cm and 1*2 cm. The room temperature is $30^°$C. Calculate the rate of heat flow through the walls of the tube. Thermal conductivity of rubber =$0*15 J/m-s—^°$C.

Solution : GIven, $\\$ $K_{rubber} = 0.15 J/m-s-^{\circ}C \qquad T_2 - T_1 = 90^{\circ}C$ $\\$ We know for radial conduction in Cylinder, $\\$ $\frac{Q}{t} = \frac{2 \pi KI(T_2 - T_1)}{In(R_2/R_1)}$ $\\$ $= \frac{2 \times 3.14 \times 15 \times 10^{-2} \times 50 \times 10^{-1} \times 90}{In(1.2/1)} = 232.5 = 233J/s$

20   A hole of radius $r_1$, is made centrally in a uniform circular disc of thickness d and radius $r_2$ . The inner surface ( a cylinder of length d and radius $r_1$,) is maintained at a temperature 8, and the outer surface (a cylinder of length d and radius $r_2)$ is maintained at a temperature $\theta_2 (\theta_1 > \theta_2)$. The thermal conductivity of the material of the disc is K. Calculate the heat flowing per unit time through the disc

Solution : $\frac{dQ}{dt} =$Rate of flow of heat $\\$Let us consider a strip at a distance r from the center of thickness dr $\\$ $\frac{dQ}{dt} = \frac{K \times 2\pi rd \times d\theta}{dr} \qquad [d\theta = Temperatue diff across the thickness dr]$ $\\$ $\Rightarrow C= \frac{K \times 2\pi rd \times d\theta}{dr} \qquad \Bigg[ c= \frac{d \theta}{dr} \Bigg]$ $\\$ $\Rightarrow C\frac{dr}{r} = K2 \pi d$ $d\theta$ $\\$ Integrating, $\\$ $\Rightarrow C\int_{r_1}^{r_2}\frac{dr}{r} = K2 \pi d \int_{\theta_1}^{\theta_2} d\theta \qquad \Rightarrow C[logr]_{r_1}^{r_2} = K2 \pi d (\theta_2 -\theta_1)$ $\\$ $C(log r_2 -logr_1) = K 2 \pi d(\theta_2 -\theta_1) \Rightarrow C\Bigg(\frac{r_2}{r_1} \Bigg) = K 2\pi d(\theta_2 -\theta_1)$ $\\$ $\Rightarrow C = \frac{K2 \pi d (\theta_2 -\theta_1)}{log(r_2/r_1)}$

21   A hollow tube has a length I, inner radius and outer radius$R_2$ The material has a thermal conductivity$K$. Find the heat flowing through the walls of the tube if (a) the flat ends are maintained at temperatures $T$, and $T _2 (T _2 > T_1)$ (b) the inside of the tube is maintained at temperature T, and the outside is maintained at $T_2$ .

Solution : $T_1 > T_2$ $\\$ $A = \pi (R_2^2 - R_1^2)$ $\\$ So, $Q = \frac{KA(T_2 - T_1)}{l} = \frac{KA(R_2^2 - R_1^2)(T_2 -T_1)}{l}$ $\\$ Considering a concentric cylindrical shell of radius 'r' and thickness 'dr'. The radial heat flow through the shell, $\\$ $H = \frac{dQ}{dt} = -KA \frac{d\theta}{dt} \qquad$[(-)ve because as r-increases $\theta$ decreases] $\\$ $A = 2 \pi rl \qquad H = -2 \pi rl k \frac{d\theta}{dt}$ $\\$ or $\int_{R_1}^{R_2}\frac{dr}{r} = -\frac{2 \pi LK}{H} \int_{T_1}^{T_2} d\theta$ $\\$ Integrating and simplifying we get,$H = \frac{dQ}{dt} = \frac{2 \pi KL(T_2 -T_1)}{log e(R_2/R_1)} = \frac{2 \pi KL(T_2 -T_1)}{In()R-2/R_1}$

22   A composite slab is prepared by pasting two plates of thicknesses $L_1$ and $L_2$ and thermal conductivities $K_1$ and $K_2$ The slabs have equal cross-sectional area. Find the equivalent conductivity of the slab.

Solution : Here the thermal conductives are in series, $\\$ $\frac{\frac{K_1A(\theta_1 -\theta_2)}{l_1} \times \frac{K_2A(\theta_1 -\theta_2)}{l_2}}{\frac{K_1A(\theta_1 -\theta_2)}{l_1} + \frac{K_2A(\theta_1 -\theta_2)}{l_2}} = \frac{KA(\theta_1 -\theta_2)}{l_1 + l_2}$ $\\$ $\Rightarrow \frac{\frac{K_1}{l_1} \times \frac{K_2}{l_2}}{\frac{K_1}{l_1} + \frac{K_2}{l_2}} = \frac{K}{l_1 + l_2}$ $\\$ $\frac{K_1K_2}{K_1l_2 + K_2l_1} = \frac{K}{l_1 + l_2} \Rightarrow K = \frac{(K_1K_2)(l_1+l_2)}{K_1l_2 + k_2l_1}$

23   Figure $(28-E2)$ shows a copper rod joined to a steel rod. The rods, have equal length and equal cross-sectional area. The free end of the copper rod is kept at $0^°$C and that of the steel rod is kept at $100^°$C. Find the temperature at the junction of the rods. Conductivity of copper = $390 W/m-^°$C and that of steel = $46 W/m-^°$C.

Solution : $K_{cu} = 390 w/m-^{\circ}C$ $\\$ Now, Since they are in series connection, $\\$ So, the heat pssed through the crosssections in the same. $\\$ So, $Q_1 =Q_2$ $\\$ Or $\frac{K_{cu} \times A \times (\theta - 0)}{l} = \frac{K_{st} \times A \times (100 - \theta)}{l}$ $\\$ $\Rightarrow 390(\theta - 0) = 46 \times 100 - 46 \theta \Rightarrow 436 \theta = 4600$ $\\$ $\Rightarrow \theta = \frac{4600}{436} = 10.55 == 10.6^{\circ}C$

24   An aluminium rod and a copper rod of equal length l'O m and cross-sectional area 1 cm 2 are welded together as shown in figure (28-E3). One end is kept at a temperature of $20^°$C and the other at $60^°$C. Calculate the amount of heat taken out per second from the hot end. Thermal conductivity of aluminium = $200 W/m-^°$C and of copper -$390 W/m-^°$C

Solution : As the Aluminium rod and Cooper rod joined are in parallel $\\$ $\frac{Q}{t} = \Bigg( \frac{Q}{t_1} \Bigg)_{Al} + \Bigg( \frac{Q}{t} \Bigg)_{Cu}$ $\\$ $\Rightarrow \frac{KA(\theta_1 - \theta_2)}{l} = \frac{K_1A(\theta_1 - \theta_2)}{l} + \frac{K_2A(\theta_1 - \theta_2)}{l}$ $\\$ $\Rightarrow K = K_1 + K_2 = (390 + 200) = 590$ $\\$ $\frac{Q}{t} = \frac{KA(\theta_1 - \theta_2)}{l} = \frac{590 \times 1 \times 10^{-4} \times (60-20)}{1} = 590 \times 10^{-4} \times 40 = 2.36$ Watt $\\$

25   Figure (28-E4) shows an aluminium rod joined to a copper rod. Each of the rods has a length of 20 cm and area of cross-section 0'20$cm^ 2$. The junction is maintained at a constant temperature $40^°$C and the two ends are maintained at $80^°$C. Calculate the amount of heat taken out from the cold junction in one minute after the steady state is reached. The conductivities are $K_{Al}= 200 W/m-^°$C and $K _{Cu} = 400 W/m-^°$C.

Solution :

$K_{Al} = 200w/m- ^{\circ}C, \qquad K_{Cu} = 400w/m- ^{\circ}C$ $A = 0.2 cm^2 = 2 \times 10^{-5} m^2$ $\\$ $l = 20cm = 2\times 10^{-1} m$ $\\$ Heat drawn per second $= Q_{Al} + Q_{Cu} = \frac{K_{Al} \times A(80-40)}{l} + \frac{K_{Cu} \times A(80-40)}{l} = \frac{2 \times 10^{-5} \times 40}{ 2 \times 10^{-1}}[200 + 400 ] = 2.4 J$ $\\$ Heat drawn per min $= 2.4 \times 60 = 144 J$

26   Consider the situation shown in figure (28-E5). The frame is made of the same material and has a uniform cross-sectional area everywhere. Calculate the amount of heat flowing per second through a cross-section of the bent part if the total heat taken out per second from the end at $100^°$C is 130 J.

Solution : $(Q/t)_{AB} = (Q/t)_{BE bent} + (Q/t)_{BE}$ $\\$ $(Q/t)_{BE bent} = \frac{KA (\theta_1 -\theta_2)}{70}, \qquad (Q/t)_{BE} = \frac{KA (\theta_1 -\theta_2)}{60}$ $\\$ $\frac{(Q/t)_{BE bent}}{(Q/t)_{BE}} = \frac{60}{70} = \frac{6}{7}$ $\\$ $(Q/t)_{BE bent} + (Q/t)_{BE} = 130$ $\\$ $\Rightarrow (Q/t)_{BE bent} +(Q/t)_{BE} 7/6= 130$ $\\$ $\Rightarrow \Bigg( \frac{7}{6} +1 \Bigg)(Q/t)_{BE bent} = 130 \qquad (Q/t)_{BE bent}= \frac{130 \times 6 }{13} = 60$

27   Suppose the bent part of the frame of the previous problem has a thermal conductivity of $780J/m-s-^°$C whereas it is $390 J/m-s-^°$C for the straight part. Calculate the ratio of the rate of heat flow through the bent part to the rate of heat flow through the straight part.

Solution : $\frac{Q}{t}bent = \frac{780 \times A \times 100}{70}$ $\\$ $\frac{Q}{t}str = \frac{390 \times A \times 100}{60}$ $\\$ $\frac{(\frac{Q}{t})bent}{(\frac{Q}{t})str} = \frac{780 \times A \times 100}{70} \times \frac{60}{390 \times A \times 100} = \frac{12}{7}$

28   A room has a window fitted with a single l'O m x 2'0 m glass of thickness 2 mm. (a) Calculate the rate of heat flow through the closed window when the temperature inside the room is 32°C and that outside is 40°C. (b) The class is now replaced by two glasspanes, each having a thickness of 1 mm and separated by a distance of 1 mm. Calculate the rate of heat flow under the same conditions of temperature. Thermal conductivity of window glass = l'O J/m-s-°C and that of air = 0'025 J/m-s-°C.

Solution : (a)$\frac{Q}{t} = \frac{KA( \theta_1 - \theta_2)}{l} = \frac{1 \times 2 \times 1(40 -32)}{2 \times 10^{-3}} =8000 J/sec$ $\\$ (b) Resistance of glass = $\frac{l}{ak_g} +\frac{l}{ak_g}$ $\\$ Resistance of air = $\frac{l}{ak_a}$ $\\$ Net resistance = $\frac{l}{ak_g} + \frac{l}{ak_g} + \frac{l}{ak_a}$ $\qquad = \frac{l}{a}\Bigg( \frac{2}{k_g} + \frac{1}{k_a} \Bigg) = \frac{l}{a}\Bigg( \frac{2k_a + k_g}{k_gk_a}\Bigg)$ $\\$ $\qquad = \frac{1 \times 10^{-3}}{2}\Bigg( \frac{2 \times 0.025 +1}{0.025} \Bigg)$ $\\$ $\frac{10^{-3} \times 1.05}{0.05}$ $\\$ $\frac{Q}{t} = \frac{\theta_1 -\theta_2}{R}= \frac{8 \times 0.05}{10^{-3} \times 1.05} = 381W$

29   The two rods shown in figure (28-E6) have identical geometrical dimensions. They are in contact with two heat baths at temperatures $100^°$C and $0^°$C. The temperature of the junction is $70^°$C. Find the temperature of the junction if the rods are interchanged.

Solution : Now; Q/t remains same in both cases, $\\$ In case I : $\frac{K_A \times A \times (100 - 70)}{l} = \frac{K_B \times A \times (70 - 0)}{l}$ $\\$ $\Rightarrow 30K_A = 70 K_B$ $\\$ In case II:$\frac{K_B \times A \times (100 - \theta)}{l} = \frac{K_A \times A \times (\theta - 0)}{l}$ $\\$ $\Rightarrow 100K_B - K_B \theta = k_A \theta$ $\\$ $\Rightarrow 100 K_B - K_B \theta = \frac{70}{30}K_B \theta$ $\\$ $100 = \frac{7}{3} \theta + \theta \qquad \Rightarrow \theta = \frac{300}{10} = 30^{\circ}C$

30   The three rods shown in figure $(28-E7)$ have identical ifometrical dimensions. Heat flows from the hot end at a rate of -$10$W in the arrangement (a). Find the rates of heat flow when the rods are joined as in arrangement (b) and in (c). Thermal conductivities of aluminium and copper are $200 W/m-^°$C and $400 W/m-^°$C respectively.

Solution : $\theta_1 - \theta_2 = 100$ $\\$ $\frac{Q}{t} = \frac{\theta_1 - \theta_2}{R}$ $\\$ $R = R_1 + R_2 + R_3 = \frac{l}{aK_{Al}} + \frac{l}{aK_{Cu}} + \frac{l}{aK_{Al}} = \frac{l}{a} \Bigg( \frac{2}{200} + \frac{1}{400} \Bigg) = \frac{l}{a}\Bigg( \frac{4+1}{400} \Bigg) = \frac{l}{a} \frac{1}{80}$ $\\$ $\frac{Q}{t} = \frac{100}{(l/a)(1/800)} \Rightarrow 40 = 80 \times 100 \times \frac{a}{l}$ $\\$ $\Rightarrow \frac{a}{l} = \frac{1}{200}$ $\\$ For (b), $\\$ $R = R_1 + R_2 = R_1 + \frac{R_{Cu}R_{Al}}{R_{Cu}+R_{Al}} = R_{Al} + \frac{R_{Cu}R_{Al}}{R_{Cu} + R_{Al}} = \frac{\frac{I}{AK_{Al}} + \frac{I}{AK_{Cu}} + \frac{I}{AK_{Al}} }{\frac{I}{A_{Al}} + \frac{I}{A_{Cu}}}$ $\\$ $= \frac{I}{AK_{Al}} + \frac{I}{A} +\frac{I}{K_{Al} + K_{Cu}} = \frac{I}{A} \Bigg( \frac{1}{200} + \frac{1}{200 + 400} \Bigg) = \frac{I}{ A} \times \frac{4}{600}$ $\\$ $\frac{Q}{t} = \frac{\theta_1 - \theta_2}{R} = \frac{100}{(I/A)(4/600)} = \frac{100 \times 600}{4} \frac{A}{I} = \frac{100 \times 600 }{4} \times \frac{1}{200} = 75$ $\\$ For (c), $\frac{1}{R} = \frac{1}{R_1} + \frac{1}{R_2} +\frac{1}{R_3} = \frac{1}{\frac{I}{aK_{Al}}} +\frac{1}{\frac{I}{aK_{Cu}}} + \frac{1}{\frac{I}{aK_{Al}}}$ $\\$ $= \frac{a}{I}(K_{Al} + K_{Cu} +K_{Al}) = \frac{a}{I}(2 \times 200 + 400) = \frac{a}{I}(800)$ $\\$ $\Rightarrow R = \frac{I}{a} + \frac{1}{800}$ $\\$ $\Rightarrow \frac{Q}{t} = \frac{\theta_1 + \theta_2}{R} = \frac{100 \times 800 \times a}{I}$ $\\$ $\frac{100 \times 800}{200} = 400W$

31   Four identical rods AB, CD, CF and DE are joined as shown in figure (28-E8). The length, cross-sectional area and thermal conductivity of each rod are l, A and K respectively. The ends A, E and F are maintained at temperatures $T_1, T_2$ and $T_3$respectively. Assuming no loss of heat to the atmosphere, find the temperature at B.

Solution : Let the temp at B be T $\\$ $\frac{Q_A}{t} = \frac{Q_B}{t} + \frac{Q_C}{t} \qquad \Rightarrow \frac{KA(T_1 - T)}{I} = \frac{KA(T - T_3)}{I + (I/2)} + \frac{KA(T - T_2 )}{I + (I/2)}$ $\\$ $\Rightarrow \frac{T_1 -T}{I } = \frac{T- T_3}{3I/2} + \frac{T- T_2}{3I/2} \qquad \Rightarrow 3T_1 - 3T = 4T - (T_2 +T_3)$ $\\$ $\Rightarrow -7T = -3T_1 - 2(T_2 + T_3) \qquad T = \frac{3T_1 + 2 (T_2 + T_3)}{7}$

32   Seven rods A, B, C, D, E, F and G are joined as shown in figure (28-E9). All the rods have equal cross-sectional area A and length I. The thermal conductivities of the rods are $K_A = K _c = K_0$, $K _B = K _D = 2K _0$ , $K_R = 3K _0$ , $K _F = 4K _0$ and $K _c = 5K _0$. The rod E is kept at a constant temperature T, and the rod G is kept at a constant temperature $T _2 (T _2 > T_1)$. (a) Show that the rod F has a uniform temperature$T = (T_1 + 2T _2 )/3$. (b) Find the rate of heat flow from the source which maintains the temperature $T _2$.

Solution :

The temp at the both ends of bar F is same $\\$ Rate of Heat flow to right = Rate of heat flow through left $\\$ $\Rightarrow (Q/t)_A + (Q/t)_C = (Q/t)_B + (Q/t)_D$ $\\$ $\Rightarrow \frac{K_A (T_1 - T) A}{I} + \frac{K_C(T_1 -T)A}{I} = \frac{K_B(T-T_2)A}{I} + \frac{K_D(T -T_2)A}{I}$ $\\$ $\Rightarrow 2K_0(T_1 -T) = 2 \times 2K_0(T -T_2)$ $\\$ $\Rightarrow T_1 - T = 2T -2T_2$ $\\$ $\Rightarrow T = \frac{T_1 + 2T_2}{3}$

33   Find the rate of heat flow through a cross-section of the rod shown in figure $(28-E10) \quad (\theta_2 > \theta_1)$. Thermal conductivity of the material of the rod is K.

Solution : $Tan \phi = \frac{r_2 -r_1}{L} = \frac{(y -r_1)}{x}$ $\\$ $\Rightarrow xr_2 -xr_1 = yL - r_1L$ $\\$ Differentiating w.r.to 'x' $\Rightarrow r_2 -r_1 = \frac{Ldy}{dx} - 0$ $\\$ $\Rightarrow \frac{dy}{dx} = \frac{r_2 - r_1}{L} \Rightarrow dx = \frac{dyL}{(r_2 - r_1)} \qquad ...(1)$ $\\$ Now, $\frac{Q}{T} = \frac{K \pi y^2 d \theta}{dx} = \frac{\theta dx}{T} = k \pi y^2 d \theta$ $\Rightarrow \frac{\theta L dy}{r_2r_1} = K \pi y^2 d \theta \qquad from (1)$ $\\$ $\Rightarrow d\theta \frac{QLdy}{(r_2 -r_1)K \pi y^2}$ $\\$ Integrating both side, $\\$ $\Rightarrow \int_{\theta_1}^{\theta_2} d\theta = \frac{QL}{(r_2 -r_1)k \pi} \int_{r_1}^{r_2} \frac{dy}{y}$ $\Rightarrow (\theta_2 - \theta_1) = \frac{QL}{(r_2 -r_1)K\pi} \times \Bigg[ \frac{-1}{y} \Bigg]_{r_1}^{r_2}$ $\\$ $\Rightarrow (\theta_2 - \theta_1) = \frac{QL}{(r_2 -r_1)K\pi} \times \Bigg[ \frac{-1}{r_1} - \frac{1}{r_2} \Bigg]$ $\\$ $\Rightarrow (\theta_2 - \theta_1) = \frac{QL}{(r_2 -r_1)K\pi} \times \Bigg[ \frac{r_2 -r_1}{r_1 +r_2} \Bigg]$ $\\$ $\Rightarrow (\theta_2 - \theta_1) = \frac{K \pi r_2 r_1 (\theta_2 -\theta_1)}{L}$

34   A rod of negligible heat capacity has length 20 cm, area of cross-section 1.0 $cm ^2$ and thermal conductivity 200 W/m-°C. The temperature of one end is maintained at 0°C and that of the other end is slowly and linearly varied from 0°C to 60°C in 10 minutes. Assuming no loss of heat through the sides, find the total heat transmitted through the rod in these 10 minutes.

Solution :

$\frac{d \theta}{ dt} = \frac{60}{10 \times 60} = 0.1^{\circ}C/sec$ $\\$ $\frac{dQ}{dt} = \frac{KA}{d} (\theta_1 - \theta_2)$ $\\$ $\qquad = \frac{KA \times 0.1}{d} + \frac{KA \times 0.2}{d} + ........ + \frac{KA \times 60}{d}$ $\\$ $\frac{KA}{d}(0.1 + 0.2 +.........+ 60) = \frac{KA}{d} \times \frac{600}{2} \times (2 \times 0.1 + 599 \times 0.1) \qquad [ \therefore a + 2a + .....+ na = n/2 {2a + (n-1)a}]$ $\\$ $= \frac{200 \times 1 \times 10^-4}{20 \times 10 ^{-2}} \times 300 \times (0.2 + 59.9 ) = \frac{200 \times 10^{-2} \times 300 \times 60.1}{20}$ $\\$ $= 3 \times 10 \times 60.1 = 1803 w$

35   A hollow metallic sphere of radius 20 cm surrounds a concentric metallic sphere of radius 5 cm. The space between the two spheres is filled with a nonmetallic material. The inner and outer spheres are maintained at 50°C and 10°C respectively and it is found that 100 J of heat passes from the inner sphere to the outer sphere per second. Find the thermal conductivity of the material between the spheres.

Solution :  $a = r_1 = 5 cm= 0.05m$ $\\$ $b = r_2 = 20 cm = 0.2 m$ $\\$ $\theta_1 = T_1 = 50 ^{\circ}C \qquad \theta_2 = T_2 = 10 ^{\circ}C$ $\\$ Now, considering a small trip of thickness 'dr' at a distance 'r' $\\$ $A = 4 \pi r^2$ $\\$ $H = -4 \pi r^2 K \frac{d \theta}{dr} \qquad$ [(-)ve because with increase of r, $\theta$decreases] $\\$ $= \int_{a}^{b}\frac{dr}{r^2} = \frac{-4 \pi K}{H} \int_{\theta_1}^{\theta_2}d\theta \qquad$ $\\$ On Integration, $H = \frac{dQ}{dt} = K \frac{4 \pi ab (\theta_1 - \theta_2)}{(b-a)}$ $\\$ Putting the values we get $\frac{K \times 4 \times 3.14 \times 5 \times 20 \times 40 \times 10^{-3}}{15 \times 10^{-2}} = 100$ $\\$ $\Rightarrow K = \frac{15}{4 \times 3.14 \times 10^{-1}} = 2.985 = 3 w/m-^{\circ}C$

36   Figure (28-Ell) shows two adiabatic vessels, each containing a mass m of water at different temperatures. The ends of a metal rod of length L, area of cross-section A and thermal conductivity K, are inserted in the water as shown in the figure. Find the time taken for the difference between the temperatures in the vessels to become half of the original value. The specific heat capacity of water is s. Neglect the heat capacity of the rod and the container and any loss of heat to the atmosphere.

Solution :

$\frac{Q}{t} = \frac{KA(T_1 - T_2)}{L} \qquad$ Rise in temp. in $T_2 \Rightarrow \frac{KA(T_1 -T_2)}{Lms}$ $\\$ Fall in Temp in $T_1 = \frac{KA(T_1-T_2)}{Lms} \qquad$ Final Temp$T_1 \Rightarrow T_1 - \frac{KA(T_1 -T_2)}{Lms}$ $\\$ Final Temp $T_2 = T_2 + \frac{KA(T_1 -T_2)}{Lms}$ $\\$ Final $\frac{\Delta T}{dt} = T_1 - \frac{KA(T_1 -T_2)}{Lms} - T_2 - \frac{KA(T_1 -T_2)}{Lms}$ $\\$ $= (T_1 -T_2) - \frac{2KA(T_1 -T_2)}{Lms} = \frac{dT}{dt} = \frac{2KA(T_1 -T_2)}{Lms} \Rightarrow \int_{(T_1 - T_2)}^{(T_1 - T_2)} \frac{dt}{(T_1 - T_2)} = \frac{-2KA}{Lms}dt$ $\\$ $\Rightarrow Ln \frac{(T_1 - T_2)/2}{(T_1 - T_2)} = \frac{-2KAt}{Lms} \qquad \Rightarrow In(1/2) = \frac{-2KAt}{Lms} \quad \Rightarrow In_2 = \frac{2KAt}{Lms} \Rightarrow t = In_2 \frac{Lms}{2KA}$

37   Two bodies of masses m, and m 2 and specific heat capacities s, and s 2 are connected by a rod of length I, cross-sectional area A, thermal conductivity K and negligible heat capacity. The whole system is thermally insulated. At time t = 0, the temperature of the first body is $T_1$, and the temperature of the second body is $T_2$ $(T_2> T_1)$. Find the temperature difference between the two bodies at time t.

Solution :

$\frac{Q}{t} = \frac{KA(T_1 -T_2)}{L} \qquad$ Rise in Temp in $T_2 \Rightarrow \frac{KA(T_1 -T_2)}{L m_1 s_1}$ $\\$ Fall in Temp in $T_1 = \frac{KA(T_1 -T_2)}{L m_2 s_2 } \qquad$ Final Temp $T_1 = T_1 - \frac{KA(T_1 -T_2)}{L m_1 s_1}$ $\\$ Final Temp $T_2 = T_2 + \frac{KA(T_1 -T_2)}{L m_1 s_1}$ $\\$ $\frac{\Delta T}{dt} = T_1 - \frac{KA(T_1 -T_2)}{L m_1 s_1} -T_2 - \frac{KA(T_1 -T_2)}{L m_2 s_2} = (T_1 -T_2) - \Bigg[ \frac{KA(T_1 -T_2)}{L m_1 s_1} + \frac{KA(T_1 -T_2)}{L m_2 s_2} \Bigg]$ $\\$ $\Rightarrow \frac{dT}{dt} = - \frac{KA(T_1 -T_2)}{L} \Bigg( \frac{1}{m_1s_1} +\frac{1}{m_2s_2} \Bigg) \qquad \Rightarrow \frac{dT}{(T_1 -T_2)} = - \frac{KA}{L} \Bigg( \frac{m_2s_2 + m_1s_1}{m_1s_1m_2s_2} \Bigg)dt$ $\\$ $\Rightarrow In \Delta t = - \frac{KA}{L} \Bigg( \frac{m_2s_2 + m_1s_1}{m_1s_1m_2s_2} \Bigg)t +C$ $\\$ At the time t = 0,$T = T_0, \qquad \Delta T = \Delta T _0 \qquad \Rightarrow C = In \Delta T_0$ $\\$ $\Rightarrow In \frac{\Delta T}{\Delta T _0} = - \frac{KA}{L} \Bigg( \frac{m_2s_2 + m_1s_1}{m_1s_1m_2s_2} \Bigg)t \quad \Rightarrow \frac{\Delta T}{\Delta T _0} = e ^ {-\frac{KA}{L} \frac{m_2s_2 + m_1s_1}{m_1s_1m_2s_2} \Bigg)t}$ $\\$ $\Rightarrow \Delta T = \Delta T_0 e ^ {-\frac{KA}{L} \frac{m_2s_2 + m_1s_1}{m_1s_1m_2s_2} \Bigg)t} = (T_2 -T_1) e ^ {-\frac{KA}{L} \frac{m_2s_2 + m_1s_1}{m_1s_1m_2s_2} \Bigg)t}$

38   An amount n (in moles) of a monatomic gas at an initial temperature $T _0$ is enclosed in a cylindrical vessel fitted with a light piston. The surrounding air has a temperature $T _S (> T _0 )$ and the atmospheric pressure is $p_a$. Heat may be conducted between the surrounding and the gas through the bottom of the cylinder. The bottom has a surface area .A, thickness x and thermal conductivity K. Assuming all changes to be slow, find the distance moved by the piston in time t.

Solution : $\frac{Q}{t} = \frac{KA(T_S - T_0)}{x} \Rightarrow \frac{nC_pdT}{dt} = \frac{KA(T_S - T_0)}{x}$ $\\$ $\Rightarrow \frac{n(5/2)RdT}{dt} = \frac{KA(T_S - T_0)}{x} \Rightarrow \frac{dT}{dt} = \frac{-2LA}{5nRx}(T_S -T_0)$ $\Rightarrow \frac{dT}{(T_S -T_0)} = -\frac{2KAdt}{5nRx} \Rightarrow In(T_S - T_0)_{T_0}^{T} = -\frac{2KAt}{5nRx}$ $\\$ $In \frac{T_S -T_0}{T_S -T_0} = -\frac{2KAdt}{5nRx} \Rightarrow T_S -T =(T_S -T_0)e^{-\frac{2KAt}{5nRx}}$ $\\$ $T = T_s -(T_s -T_0)e^{-\frac{2KAt}{5nRx}} = T_S + (T_S +T_0)e^{-\frac{2KAt}{5nRx}}$ $\\$ $\Delta T = T -T_0 = (T_S -T_0 ) + (T_S -T_0) e^{-\frac{2KAt}{5nRx}} = (T_S -T_0) + \Bigg( 1 + e^{-\frac{2KAt}{5nRx}} \Bigg)$ $\\$ $\Rightarrow \frac{P_a AL}{nR} = (T_S -T_0) + \Bigg( 1 + e^{\frac{2KAt}{5nRx}} \Bigg) \qquad$ [$p_adv = nRdt \quad P_aAl = nRdt \quad dT = \frac{P_aAl}{nR}$] $\\$ $L = \frac{nR}{P_aA}(T_S -T_0) + \Bigg( 1-e^{-\frac{2KAt}{5nRx}} \Bigg)$

39   Assume that the total surface area of a human body is $1.6 m ^2$ and that it radiates like an ideal radiator. Calculate the amount of energy radiated per second by the body if the body temperature is 37°C. Stefan constant o is $6.0 \times 10 ^{-3} W/m ^2 -K ^4$

Solution :

$A = 1.6 m^2, \qquad T = 37^{\circ}C = 310 K, \qquad \sigma = 6.0 \times10 ^{-8} W/m ^2 -K ^4$ $\\$ Energy radiated per second $= A \sigma T^4 = 1.6 \times 6 \times 10 ^{-8} \times (310)^4 = 8865801 \times 10^{-4} = 886.58 = 887 J$

40   Calculate the amount of heat radiated per second by a body of surface area 12 cm kept in thermal equilibrium in a room at temperature 20°C. The emissivity of the surface - 0.80 and o - $6.0 \times 10 ^{-8 }W/m^ 2 -K ^4$

Solution :

$A = 12 cm^2 = 12 \times 10^{-4} m^2 \qquad T = 20^{\circ} = 293 K$ $\\$ $e = 0.8 \qquad \sigma = 6 \times 10^{-8} w/m^2 -k^4$ $\\$ $\frac{Q}{t} = Ae \sigma T^4 = 12 \times 10^{-4} \times 0.8 \times 6 \times 10^{-8}(293^4) = 4.245 \times 10^{12} \times 10^{-13} = 0.4245 = 0.42$

41   A solid aluminium sphere and a solid copper sphere of twice the radius are heated to the same temperature and are allowed to cool under identical surrounding temperatures. Assume that the emissivity of both the spheres is the same. Find the ratio of (a) the rate of heat loss from the aluminium sphere to the rate of heat loss from the copper sphere and (b) the rate of fall of temperature of the aluminium sphere to the rate of fall of temperature of the copper sphere. The specific heat capacity of aluminium = 900 J/Kg-°C and that of copper - 390 J/Kg-°C. The density of copper = 3'4 times the -density of aluminium

Solution :

$E \rightarrow$Energy radiated per unit time $\\$ Rate of heat flow $\rightarrow$ Energy radiated. $\\$ (a) Per time = $E \times A$ $\\$ So, $E_{Al} = \frac{e \sigma T^4 \times A}{ e \sigma T^4 \times A} = \frac{4 \pi r^2}{4 \pi (2r)^2} = \frac{1}{4} \qquad \therefore 1:4$ $\\$ (b) Emissivity of both are same $= \frac{m_1S_1dT_1}{m_2S_2dT_2} = 1$ $\\$ $\Rightarrow \frac{dT_1}{dT_2} = \frac{m_2S_2}{m_1S_1} = \frac{s_14 \pi r_1^3 \times S_2}{s_24 \pi r_2^3 \times S_1} = \frac{1 \times \pi \times 900}{ 3.4 \times 8 \pi \times 390} = 1: 2 : 9$

42   A 100 W bulb has tungsten filament of total length l'O m and radius $4 \times 10 ^{-5}$ m. The emissivity of the filament is 0.8 and $\sigma - 6.0 \times 10^ {-8} W/m^2 -K^4$ Calculate the temperature of the filament when the bulb is operating at correct wattage.

Solution :

$\frac{Q}{t} = Ar \sigma T^4$ $\\$ $\Rightarrow T^4 = \frac{\theta}{teA \sigma } \Rightarrow T^4 = \frac{100}{0.8 \times 2 \times 3.14 \times 4 \times 10^{-5} \times 1 \times 6 \times 10^{-8}}$ $\\$ $\Rightarrow T = 1697.0 = 1700 K$

43   A spherical ball of surface area $20 cm^ 2$ absorbs any radiation that falls on it. It is suspended in a closed box maintained at 57°C. (a) Find the amount of radiation falling on the ball per second, (b) Find the net rate of heat flow to or from the ball at an instant when its temperature is 200°C. Stefan constant = $6.0 \times 10^{-8} W/m^2 -K^4$

Solution :

$(a) \quad A = 20cm^2 = 20 \times 10^{-4} m^2$ $\\$ $E = A \sigma T^4 = 20 \times 10^{-4} \times 6 \times 10^{-8} \times (330)^4 \times 10^4 = 1.42 J$ $\\$ $(b) \quad \frac{E}{t} = A \sigma e (T_1^4 - T_2^4), \qquad A = 20cm^2 = 20 \times 10^{-4} m^2$ $\\$ $\sigma = 6 \times 10^{-8}, \qquad T_1 = 473K, \qquad T_2 = 330 K$ $\\$ $=20 \times 10^{-4} \times 6 \times 10^{-8} \times 1[(473)^4- (330)^4]$ $\\$ $=20 \times 6 \times [5.005 \times 10^{10} - 1.185 \times 10^{10}]$ $\\$ $=20 \times 6 \times 3.82 \times 10^{-2} = 4.58w \qquad$ from the ball.

44   A spherical tungsten piece of radius $1.0$cm is suspended in an evacuated chamber maintained at $300$K. The piece is maintained at $1000$ K by heating it electrically. Find the rate at which the electrical energy must be supplied. The emissivity of tungsten is $0.30$ and the Stefan constant o is $6.0 x 10 ^{-8} W/m^2 - K^{4}$ .

Solution :

$r = 1cm = 1 \times 10^{-3}m$ $\\$ $A = 4 \pi (10^{-2})^2 = 4 \pi \times 10^{-4} m^2$ $\\$ $E = 0.3, \qquad \sigma = 6 \times 10^{-8}$ $\\$ $\frac{E}{t} = A \sigma e (T_1^4 - T_2^4)$ $\\$ $=0.3 \times 6 \times 10^{-8} \times 4 \pi \times 10^{-4} \times [(100)^4 - (300)^4]$ $\\$ $= 0.3 \times 6 \times 4 \pi \times 10 ^{-12} \times [1 - 0.0081] \times 10^{12}$ $\\$ $= 0.3 \times 6 \times 4 \times 3.14 \times 9919 \times 10^{-4}$ $\\$ $= 4 \times 18 \times 3.14 \times 9919 \times 10^{-5} = 22.4 == 22W$

45   A cubical block of mass 1.0 kg and edge 5.0 cm is heated to 227°C. It is kept in an evacuated chamber maintained at 27°C. Assuming that the block emits radiation like a blackbody, find the rate at which the temperature of the block will decrease. Specific heat capacity of the material of the block is 400 J/kg-K.

Solution : Since the Cube can be assumed as black body $\\$ $e = l$ $\\$ $\sigma = 6 \times 10^{-8} w/m^2-K^4$ $\\$ $A = 6 \times 25 \times 10^{-4} m^2$ $\\$ $m = 1 Kg$ $\\$ $s = 400J/Kg-^{\circ}K$ $\\$ $T_1 = 227^{\circ} C = 500 K$ $\\$ $T_2 = 27^{\circ} C = 300 K$ $\\$ $\Rightarrow ms \frac{d \theta}{dt} = e \sigma A(T_1^4 - T_2^4)$ $\\$ $Rightarrow \frac{d \theta }{dt} = \frac{e \sigma A(T_1^4 - T_2^4)}{ms}$ $\\$ $= \frac{1 \times 6 \times 10^{-8} \times 6 \times 25 \times 10^{-4} \times [(500)^4 - (300)^4] }{1 \times 400}$ $\\$ $= \frac{36 \times 25 \times 544 }{400} \times 10^{-4} = 1224 \times 10^{-4} = 0.1224 ^{\circ}C/s$

46   A copper sphere is suspended in an evacuated chamber maintained at 300 K. The sphere is maintained at a constant temperature of 500 K by heating it electrically. A total of 210 W of electric power is needed to do it. When the surface of the copper sphere is completely blackened, 700 W is needed to maintain the same temperature of the sphere. Calculate the emissivity of copper.

Solution :

$Q = e \sigma A (T_2^4 - T_1^4)$ $\\$ For any body, $210 = e A \sigma [(500)^4-(300)^4]$ $\\$ For black body,$700 = 1 \times A \sigma [(500)^4 -(300)^4]$ $\\$ Dividing $\frac{210}{700} = \frac{e}{1} \Rightarrow e =0.3$

47   A spherical ball A of surface area $20 cm^2$ is kept at the centre of a hollow spherical shell B of area $80 cm^2$ . The surface of A and the inner surface of B emit as blackbodies. Assume that the thermal conductivity of the material of B is very poor and that of A is very high and that the air between A and B has been pumped out. The heat capacities of A and B are 42J/°C and 82 J/°C respectively. Initially, the temperature of A is 100°C and that of B is 20°C. Find the rate of change of temperature of A and that of B at this instant. Explain the effects of the assumptions listed in the problem.

Solution : $A_A = 20 cm^2, \qquad A_B = 80cm^2$ $\\$ $(mS)_A = 42J/^{\circ}C, \qquad (mS)_B = 82 J/^{\circ}C$ $\\$ $T_A = 100^{\circ}C, \qquad T_B = 20^{\circ}C$ $\\$ $K_B$ is low thust it is poor conducter and $K_A$ is high $\\$ Thus A will absorb no heat and conduct all $\\$ $\Bigg( \frac{E}{t} \Bigg)_A = \sigma A_A [ (373)^4 - (293)^4 ] \Rightarrow (mS)_A \Bigg( \frac{d \theta}{dt} \Bigg)_A = \sigma A_A [ (373)^4 - (293)^4 ]$ $\\$ Similarly $\Bigg(\frac{d \theta}{dt} \Bigg)_B = 0.043^{\circ}C/S$

48   A cylindrical rod of length 50 cm and cross-sectional area $1 cm^2$ is fitted between a large ice chamber at 0°C and an evacuated chamber maintained at 27°C as shown in figure (28-E12). Only small portions of the rod are inside the chambers and the rest is thermally insulated from the surrounding. The cross-section going into the evacuated chamber is blackened so that it completely absorbs any radiation falling on it. The temperature of the blackened end is 17°C when steady state is reached. Stefan constant $\sigma = 6 \times 10^{-8} W/m^2 - K^4$ . Find the thermal conductivity of the material of the rod.

Solution :

$\frac{Q}{t} = eAe(T_2^4 - T_1^4)$ $\\$ $\Rightarrow \frac{Q}{At} = 1 \times 6 \times 10^{-8}[(300)^4 - (290)^4] \qquad = 6 \times 10^{-8} (81 \times 10^{8} - 70.7 \times 10^{8}) = 6 \times 10.3$ $\\$ $\frac{Q}{t} = \frac{KA(\theta_1 - \theta_2)}{I}$ $\\$ $\Rightarrow \frac{Q}{tA} = \frac{KA(\theta_1 - \theta_2)}{I} = \frac{K \times 17 }{0.5} = 6 \times 10.3 = \frac{K \times 17}{0.5} \Rightarrow K = \frac{6 \times 10.3 \times 0.5 }{17} = 1.8$

49   One end of a rod of length 20 cm is inserted in a furnace at 800 K. The sides of the rod are covered with an insulating material and the other end emits radiation like a blackbody. The temperature of this end is 750 K in the steady state. The temperature of the surrounding air is 300 K. Assuming radiation to be the only important mode of energy transfer between the surrounding and the open end of the rod, find the thermal conductivity of the rod. Stefan constant $\sigma = 6.0 \times 10 ^{-8} W/m ^2 -K ^4$.

Solution : $\sigma = 6 \times ^{-8}w/m^2- k^4$ $\\$ $L = 20cm = 0.2 m, \qquad K =?$ $\\$ $\Rightarrow E = \frac{KA (\theta_1 - \theta_2)}{d} = A \sigma (T_1^4 - T_2^4)$ $\\$ $\Rightarrow K = \frac{s(T_1 -T_2) \times d}{\theta_1 - \theta_2} = \frac{6 \times 10^{-8} \times (750^4 - 300^4 ) \times 2 \times 10^{-1}}{50}$ $\\$ $\Rightarrow K = 73.993 = 74$

50   A calorimeter of negligible heat capacity contains 100 cc of water at 40°C. The water cools to 35°C in 5 minutes. The water is now replaced by K-oil of equal volume at 40°C. Find the time taken for the temperature to become 35°C under similar conditions. Specific heat capacities of water and K-oil are 4200 J/Kg-K and 2100 J/Kg-K respectively. Density of K-oil -$800 kg/m^3$

Solution :

$v = 100cc$ $\\$ $\Delta \theta = 5^{\circ}C$ $\\$ $t = 5 min$ $\\$ For Water $\frac{mS \Delta \theta}{dt} = \frac{KA}{I} \Delta \theta$ $\\$ $\Rightarrow \frac{100 \times 10^{-3} \times 1000 \times 4200}{5} = \frac{KA}{I}$ $\\$ For Kerosene, $\frac{ms}{at} = \frac{KA}{I}$ $\\$ $\Rightarrow \frac{100 \times 10 ^{-3} \times 800 \times 2100}{t} = \frac{KA}{I}$ $\\$ $\Rightarrow \frac{100 \times 10^{-3} \times 800 \times 2100}{t} = \frac{100 \times 10^{-3} \times 1000 \times 4200}{ 5}$ $\\$ $\Rightarrow T = \frac{5 \times 800 \times 2100}{1000 \times 4200} = 2 min$

51   A body cools down from 50°C to 45°C in 5 minutes and to 40°C in another 8 minutes. Find the temperature of the surrounding.

Solution :

$50^{\circ}C \qquad 45^{\circ}C \qquad 40^{\circ}C$ $\\$ Let the surrounding temperature be $'T'^{\circ}C$ $\\$ Avg. $t = \frac{50 + 45}{2} = 47.5$ $\\$ Avg. temperature differrence from surrounding, $\\$ $T =47.5 - T$ $\\$ Rate of fall of temp = $\frac{50-45}{5} = 1^{\circ}C/mm$ $\\$ From Newton's Law $1^{\circ}C/mm = bA \times t$ $\\$ $\Rightarrow bA = \frac{1}{t} = \frac{1}{47.5 - T}$ $\\$ In second case, $\\$ Avg, temp $= \frac{40 + 45}{2} = 42.5$ $\\$ Avg. temp. diff. from surrounding $t' = 42.5 -t$ $\\$ Rate of fall of temp = $\frac{45 -40 }{8} = \frac{5}{8} ^{\circ}/mm$ $\\$ From Newton's Law, $\\$ $\frac{5}{B} = bAt'$ $\\$ $\Rightarrow \frac{5}{8} = \frac{1}{(47.5 -T)} \times (42.5 - T)$ $\\$ By C & D [Componendo & Dividendo method] $\\$ We find, $T = 34.1^°C$

52   A calorimeter contains 50 g of water at 50°C. The temperature falls to 45°C in 10 minutes. When the calorimeter contains 100 g of water at 50°C, it takes 18 minutes for the temperature to become 45°C. Find the water equivalent of the calorimeter.

Solution :

Let the water eq. of caloriemeter = m $\\$ $\frac{(m+50 \times 10^{-3}) \times 4200 \times 5}{10} =$ Rate of heat flow $\\$ $\frac{(m+50 \times 10^{-3}) \times 4200 \times 5}{18} =$ Rate of flow $\\$ $\Rightarrow \frac{(m+50 \times 10^{-3}) \times 4200 \times 5}{10} = \frac{(m+50 \times 10^{-3}) \times 4200 \times 5}{18}$ $\\$ $\Rightarrow (m+50 \times 10^{-3})18 = 10m + 1000 \times 10^{-3}$ $\\$ $\Rightarrow 18m +18 \times 50 \times 10^{-3} = 10m + 1000 \times 10^{-3}$ $\\$ $\Rightarrow 8m = 100 \times 10^{-3} Kg$ $\\$ $\Rightarrow m = 12.5 \times 10^{-3}kg = 12.5g$

53   A metal ball of mass 1 kg is heated by means of a 20 W heater in a room at 20°C. The temperature of the ball becomes steady at 50°C. (a) Find the rate of loss of heat to the surrounding when the ball is at 50°C. fa) Assuming Newton's law of cooling, calculate the rate of loss of heat to the surrounding when the ball is at 30°C. (c) Assume that the temperature of the ball rises uniformly from 20°C to 30°C in 5 minutes. Find the total loss of heat to the surrounding during this period, (d) Calculate the specific heat capacity of the metal

Solution : In steady state condition as no heat is absorbed, the rate of loss of heat by conduction is equal to that of the supplied i.e. $H = P$ $\\$ $m = 1Kg,$ Power of Heater = 20W, Room Temp= $20^{\circ}C$ $\\$ (a) $H= \frac{d \theta }{dt} = P = 20 Watt$ $\\$ (b) by Newton's Law of cooling, $\frac{-d \theta}{dt} = K(\theta - \theta_0)$ $\\$ $-20 = K(50-20) \Rightarrow K = \frac{2}{3}$ $\\$ Again, $\frac{-d \theta}{dt} = K(\theta - \theta_0) = \frac{2}{3} \times (30-20) \frac{20}{3}$ $\\$ (c) $\Bigg( \frac{dQ}{dt} \Bigg)_{20} = 0, \qquad \Bigg( \frac{dQ}{dt} \Bigg)_{30} = \frac{20}{3}, \qquad \Bigg( \frac{dQ}{dt} \Bigg)_{avg} = \frac{10}{3}$ $\\$ $T = 5 min = 300$ $\\$ Heat liberated $= \frac{10}{3} \times 300 = 1000J$ $\\$ Now Heat absor bed = Heat Supplied - Heat Radiated = $6000 - 1000=5000 J$ $\\$ Now, $m \Delta \theta^{'} = 5000$ $\\$ $S = \frac{5000}{m \Delta \theta}$= $\frac{5000}{1 \times 10}$= 500 J Kg$^{-1}$ $^{\circ}$ C$^{-1}$

54   A metal block of heat capacity 80 J/°C placed in a room at 20°C is heated electrically. The heater is switched off when the temperature reaches 30°C. The temperature of the block rises at the rate of 2 °C/s just after the heater is switched on and falls at the rate of 0'2 °C/s just after the heater is switched off. Assume Newton's law of cooling to hold, (a) Find the power of the heater, (b) Find the power radiated by the block just after the heater is switched off. (c) Find the power radiated by the block when the temperature of the block is 25°C. (d) Assuming that the power radiated at 25°C respresents the average value in the heating process, find the time for which the heater was kept on.

Solution :

Given: $\\$ Heat capacity = $m \times s = 80J/^{\circ}C$ $\\$ $\Bigg( \frac{d \theta}{dt} \Bigg)_{increase} = 2^{\circ}C/s$ $\\$ $\Bigg( \frac{d \theta}{dt} \Bigg)_{decrease} = 0.2^{\circ}C/s$ $\\$ (a) Power of Heater = $mS\Bigg( \frac{d \theta}{dt} \Bigg)_{increasing} = 80 \times 2 = 160W$ $\\$ (b) Power radiated = $mS\Bigg( \frac{d \theta}{dt} \Bigg)_{decreasing} = 80 \times 0.2 = 16W$ $\\$ (c) Now $mS \Bigg( \frac{d \theta}{dt} \Bigg)_{decreasing} = K(T -T_0)$ $\\$ $\Rightarrow 16 = K (30-20) \qquad \Rightarrow K = \frac{16}{10} = 1.6$ $\\$ Now, $\frac{d \theta }{dt} = K(T-T_0)= 1.6 \times (30-25) = 1.6 \times 5 = 8W$ $\\$(d) P.t = $H \Rightarrow 8 \times t$

55   A hot body placed in a surrounding of temperature $\theta_0$ obeys Newton's law of cooling $\frac{d \theta}{dt} = - k(\theta - \theta_0)$. Its temperature at t = 0 is $\theta_1$. The specific heat capacity of the body is s and its mass is rn. Find (a) the maximum heat that the body can lose and fa) the time starting from t = 0 in which it will lose 90% of this maximum heat.

Solution :

$\frac{d \theta}{dt} = -K(T -T_0)$ $\\$ Temp at $t= 0$ is $\theta_1$ $\\$ (a) Max Heat that the body can loose = $\Delta Q_m = ms(\theta_1 - \theta_0) \qquad (\therefore as, \Delta t = \theta_1 - \theta_0 )$ $\\$ (b) if the body loses $90\%$ of the max heat the decrease in its temp will be $\\$ $\frac{\Delta Q_m \times 9}{10ms} = \frac{(\theta_1 - \theta_0) \times 9}{10}$ $\\$ If it takes time $t_1$, for this purpose, the temp. at $t_1$, $\\$ $= \theta_1 - (\theta_1 - \theta_0) \frac{9}{10} = \frac{10 \theta_1 - 9 \theta_1 - 9\theta_0}{10} = \frac{\theta_1 - 9\theta_0}{10} \times 1$ $\\$ Now, $\frac{d \theta}{dt} = -K(\theta - \theta_1)$ $\\$ Let $\theta = \theta_1$ at t = 0; & $\theta$ be temp. at time t $\\$ $\int_{\theta}^{\theta} \frac{d \theta}{\theta - \theta_0} = - K\int_{0}^{1}dt$ $\\$ or, $In\frac{\theta - \theta_0}{\theta_1 - \theta_0} = -Kt$ $\\$ or, $\theta - \theta_0 = (\theta_1 - \theta_0)e^{-Kt} \qquad ......(2)$ $\\$ Putting value in the Eq(1) and Eq(2), $\\$ $\frac{\theta_1 - 9\theta_0}{10} - \theta_0 = (\theta_1 -\theta_0)e^{-Kt}$ $\\$ $\Rightarrow t_1 = \frac{In10}{K}$