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General Physics I:

Section 10.4 Examples A

Temperature & Expansion.

Example 10.4.1.

Nitrogen liquefies at 63 K. What is this temperature in
  1. degrees Celsius?
  2. degrees Fahrenheit?
Solution.
Given: \(T = 63 K \)
  1. \begin{equation*} T_{C} = T - 273^{o}C, T_{C} = -210^{o}C \end{equation*}
  2. \begin{equation*} T_{F} = T_{C} + 32^{o}F, T_{F} = -346^{o}F \end{equation*}

Example 10.4.2.

A steel rail 50 m long and cross sectional area \(50 cm^{2}\) is laid out on a hot summer day when the temperature is \(35^{o}C\text{.}\) The next winter the temperature drops to \(20^{o}C\) below zero. How great a force would be required to keep the ends of the rail from moving? For steel, \(\alpha = 11\times10^{-6}/^{o}C\text{,}\) \(\rho = 7.86\times 10^{3} \,kg/m^{3},\) \(Y = 20\times 10^{10} \,N/m^{2}\text{.}\)
Solution.
Given: \(l = 50 \,m, \quad A = 5.0\times 10^{-5} \,m^{2}, \quad T_{o} = 35^{o}C,\) \(T = -20^{o}C, \quad\alpha = 11\times 10^{-6}/^{o}C. \)
\begin{equation*} \Delta l = \alpha l_{o} \Delta T, \end{equation*}
\begin{equation*} \Delta T = T - T_{o} = -20^{o}C-35^{o}C= -55^{o}C \end{equation*}
\begin{equation*} \therefore\quad \Delta l = 11\times 10^{-6}(/^{o}C)\times50 \,m\times(-55^{o}C) = -3.025\times 10^{-2} \,m, \end{equation*}
\begin{equation*} \sigma = \frac{F}{A}, \end{equation*}
\begin{equation*} \epsilon = \frac{\Delta l}{l_{o}} = 6.05\times 10^{-4}, \end{equation*}
\begin{equation*} \quad Y = \frac{\sigma}{\epsilon}=\frac{F}{A\epsilon} \end{equation*}
\begin{equation*} F = YA\epsilon = 20\times 10^{8} \,Pa\times 55\times 10^{-4}\,m^{2}\times 6.05\times 10^{-4}=6655 \,N \end{equation*}

Example 10.4.3.

A sheet of aluminum 2 m long, 1 m wide, and 1 mm thick at \(25^{o}C\) is heated to \(50^{o}C\text{.}\) If \(\alpha_{al} = 24\times 10^{-6}/^{o}C\text{,}\) how much does the area of the sheet increase?
Solution.
Given: \(l = 2 \,m,\quad w = 1 \,m, \quad h = 0.001 \,m,\) \(T_{o} = 25^{o}C, \quad T = 50^{o}C, \quad \alpha = 24\times 10^{-6} /^{o}C \)
\begin{equation*} \Delta T = T - T_{o} =50-25= 25^{o}C, \end{equation*}
\begin{equation*} A_{o} = lw = 2\times 1= 2 \,m^{2} \end{equation*}
\begin{equation*} \therefore\quad \Delta A = \beta A_{o} \Delta T =2 \alpha A_{o} \Delta T \end{equation*}
\begin{equation*} = 2\times 24\times 10^{-6} /^{o}C \times2 m^{2}\times 25^{o}C = 2.4\times 10^{-3} m^{2} \end{equation*}

Example 10.4.4.

A 1 L glass flask \((\alpha = 9\times 10^{-6}/^{o}C)\) is filled to the mark with ethyl alcohol \((\gamma = 1.12\times 10^{-4}/^{o}C)\) at \(0^{o}C\text{.}\) If the system is heated to \(50^{o}C\text{,}\)
  1. how much does the glass flask expand?
  2. How much does the alcohol expand?
  3. How high does the alcohol flow up the neck of the flask if its cross-sectional area is \(1 \,cm^{2}\text{?}\)
Solution.
Given: \(V_{o} = 1.0\times 10^{-3} \,m^{3}, \quad \alpha = 9\times 10^{-6} /^{o}C,\) \(\gamma_{a} = 1.12\times 10^{-4}/^{o}C.\) \(\Delta T = 50^{o}C,\) \(A_{o}=1 \,cm^{2}=1\times 10^{-4} \,m^{2},\)
\begin{equation*} \gamma_{g} = 3 \alpha \end{equation*}
  1. \begin{equation*} \therefore\quad \Delta V_{g} = \gamma_{g} V_{o} \Delta T = 3\alpha V_{o} \Delta T= 1.35\times 10^{-6} \,m^{3} \end{equation*}
  2. \begin{equation*} \Delta V_{a} = \gamma_{a}V_{o} \Delta T = 1.12\times 10^{-4} \times 1.0\times 10^{-3} \times 50 = 5.6\times 10^{-6} \,m^{3} \end{equation*}
  3. \begin{equation*} \Delta V' = \Delta V_{a}-\Delta V_{g} = 5.6\times 10^{-6}-1.35\times 10^{-6} = 4.25\times 10^{-6} \,m^{3} \end{equation*}
    \begin{equation*} \text{But,}\quad \Delta V' = A_{o} h \end{equation*}
    \begin{equation*} \Rightarrow\quad h=\frac{\Delta V'}{A_{o}} =\frac{4.25\times 10^{-6}}{1\times 10^{-4}}=4.25\times 10^{-2} \,m \end{equation*}

Calorimetry.

Example 10.4.5.

How much heat is required to convert a 100 g ice cube taken from the refrigerator at \(-10^{o}C\) into super heated steam at \(110^{o}C\) if the conversion is performed at atmospheric pressure? Note: c(ice) = c(steam) = \(0.5 \, cal/g^{o}C,\) \(L_{f}(water) = 80 \,cal/g,\) \(L_{v}(water) = 540 \,cal/g.\)
Solution.
Given: \(m_{ice}=100\,g, \quad T_{1} =-10^{o}C, \quad m_{steam}=100\,g, \quad T_{2} =110^{o}C.\)
\begin{equation*} Q= Q_{ice}+Q_{ice \to water}+Q_{water}+Q_{water \to steam}+Q_{steam} \end{equation*}
\begin{align*} Q \amp= mc(T_{f}-T_{i})_{ice}+(m L_{f})_{ice\to water} \\ \amp +mc(T_{f}-T_{i})_{water}+(m L_{v})_{water\to steam}+mc(T_{f}-T_{i})_{steam} \end{align*}
\begin{align*} Q\amp = m\left[c(T_{f}-T_{i})_{ice}+( L_{f})_{ice\to water}\right.\\ \amp \left.+c(T_{f}-T_{i})_{water}+( L_{v})_{water\to steam}+c(T_{f}-T_{i})_{steam}\right] \\ \text{or,}\quad Q \amp = 100 \left[(0.5)(0+10)+ 80+1(100-0)+ 540+0.5(110-100)\right] \\ \amp =73000 \,cal. \end{align*}

Example 10.4.6.

A 400 g piece of iron is taken from an oven at \(300^{o}C\) and placed in a 100 g aluminum calorimeter containing 100 g of ice at \(0^{o}C\text{.}\)
  1. How much ice melts?
  2. What is the final temperature of the system? Note: \(c(irn) = 470 \,J/kg.K, c(aluminum) = 910 \,J/kg.K,\) \(L_{f}(w) = 334 \,J/g,\) \(c(w) = 4186 \,J/kg.K\text{.}\)
Solution.
Given: Mass of aluminum calorimeter, \(m_{a} = 100 \,g = 0.1 \,kg,\) mass of ice, \(m_{ice} = 100 \,g = 0.1 \,kg,\) and mass of iron, \(m_{ir} = 400 \,g = 0.4 \,kg.\) Initial temperature of ice and calorimeter, \(T_{1}=0^{o}C,\) initial temperature of iron, \(T_{2}=300^{o}C. \) When hot iron is placed in calorimeter iron losses its heat and calorimeter and contents gains heat till the final temperature of the mixture reaches to equilibrium at \(T_{f}.\)
Now first calculate amount of heat needed for ice to melt completely,
\begin{equation*} m_{ice}L_{ice} = 0.1\,kg(334 \,J/g)(1000\,g/1\,kg) = 33400 \,J \end{equation*}
Heat lost by iron if cool to \(0^{o}C\) = \(m_{ir}c_{ir}(T_{2}-T_{f}) = 0.4 \,kg (470\,J/kgK) (300-0) = 56400 \,J.\) Hence more heat is available in container than ice requires to melt completely.
  1. All ice melts, i.e., \(m=100 \,g.\)
  2. Now from principle of calorimetry, Heat loss by iron = heat gained by calorimeter + ice.
    \begin{equation*} m_{ir}c_{ir}(T_{2}-T_{f}) = \left[mL_{f}(ice)+mc(T_{f}-T_{1})_{w}+m_{a}c_{a}(T_{f}-T_{1})\right] \end{equation*}
    \begin{equation*} or, \, 0.4\,kg(470 \,J/kg.K)(300-T_{f}) = \left[0.1\,kg(334\,J/g)(1000\,g/1\,kg)\right. \end{equation*}
    \begin{equation*} \left.+0.1(4186)(T_{f}-0)+0.1\,kg (910 \,J/kg.K)\right](T_{f}-0) \end{equation*}
    \begin{equation*} or,\, 188\times 300-188 T_{f} = 33400+418.6 T_{f}+91(T_{f}) \end{equation*}
    \begin{equation*} or,\, 56400-33400 = (91+188+418.6) T_{f} \end{equation*}
    \begin{equation*} \therefore\quad T_{f} = \frac{23000}{697.6} =32.97^{o}C \end{equation*}

Example 10.4.7.

Table salt has the chemical formula of \(NaCl\) and a specific heat capacity of \(879 \,J/kg.K.\) The atomic mass of sodium is 22.99 and of chlorine is 35.453.
  1. What is the molar mass of salt?
  2. What is the molar heat capacity of salt?
  3. How many joules of heat are required to heat 2 moles of salt from room temperature \((25^{o}C)\) to boiling temperature \((100^{o}C)\text{?}\)
Solution.
Given: \(c=879 \,J/kg.K,\) \(M_{Na}=22.99 \,g/mol,\) \(M_{Cl}=35.453 \,g/mol,\quad n=2 \,mol, \) \(T_{1}= 25^{o}C,\quad T_{2}=100^{o}C.\)
  1. Molecular mass
    \begin{equation*} M=M_{Na}+M_{Cl} \end{equation*}
    \begin{equation*} = (22.99 +35.453) \,g/mol = 58.443 \,g/mol = 0.05844 \,kg/mol \end{equation*}
  2. Molar heat capacity,
    \begin{equation*} C=Mc=(0.05844 \,kg/mol)\times(879 \,J/kg K) =51.4 \,J/mol.K \end{equation*}
  3. \begin{equation*} Q=nC\Delta T = nC\left(T_{2}-T_{1}\right) = 2\times 51.4\times(100-25) = 7710 \,J \end{equation*}

Example 10.4.8.

A lead bullet of mass 10 g, traveling at 500 m/s, strikes a target and comes to rest. The specific heat of solid lead is \(128 \,J/kg.^{o}C\text{,}\) lead melts at \(327.3^{o}C\text{,}\) and its heat of fusion is \(2.45\times 10^{4} \,J/kg\text{.}\) If the bullet is at \(30^{o}C\) before hitting the target and all the heat generated by the collision goes into the bullet, what is the final temperature of the bullet?
Solution.
Given: \(m=10 \,g =0.01\,kg, \quad v_{1}=500 \,m/s,\quad v_{2}=0,\) \(c_{Pb} = 128 \,J/kg.^{o}C,\) \(T_{1} = 327.3^{o}C, \quad T_{o}=30^{o}C, \) \(L_{f} = 2.45\times 10^{4} \,J/kg, T_{f} =?\)
Heat produced by bullet = change in kinetic energy of bullet.
\begin{equation*} Q=E_{k} \end{equation*}
\begin{equation*} or,\quad mc(T_{1}-T_{o})+mL_{f}+mc(T_{f}-T_{o})=\frac{1}{2}m(v_{1}^{2}-v_{2}^{2}) \end{equation*}
\begin{equation*} \text{or,}\quad c(T_{1}-T_{o})+L_{f}+c(T_{f}-T_{1})=\frac{1}{2}(v_{1}^{2}) \end{equation*}
\begin{equation*} \text{or,}\quad 128 (327.3-30) +2.45\times 10^{4} + 128 (T_{f}-327.3)= \frac{1}{2}(500^{2}) \end{equation*}
\begin{equation*} \text{or,}\quad -3840 +2.45\times 10^{4} + 128 T_{f}= 125000 \end{equation*}
\begin{equation*} \text{or,}\quad 128 T_{f}= 125000+3840-2.45\times 10^{4} = 104340 \end{equation*}
\begin{equation*} \qquad \therefore \quad T_{f} = 815^{o}C \end{equation*}

Example 10.4.9.

A piece of iron (\(c=0.107 \,cal/g^{o}C\)) having a mass of 300 g and temperature of \(200^{o}C\) is placed in a wall insulated and massless calorimeter containing 500 g water (\(c=1 \,cal/g^{o}C\)) at \(30^{o}C.\)
  1. What is the final temperature of the system?
  2. What is the change in entropy of the iron?
  3. What is the change in entropy of the water?
  4. What is the change in entropy of the universe?
  5. Is this result consistent with the second law of thermodynamics. Why?
Solution.
Given: \(m_{i}=300 \,g, \) \(T_{h}=200^{o}C,\) \(m_{w}=500 \,g,\) \(T_{c}=30^{o}C.\)
  1. Heat loss = heat gain
    \begin{equation*} m_{i}c_{i}(T_{h}-T_{f}) = m_{w}c_{w}(T_{f}-T_{c}) \end{equation*}
    \begin{equation*} \text{or,}\quad (300 \,g) (0.107 \,cal/g^{o}C)(200-T) = (500 \,g) (1 \,cal/g^{o}C)(T-30) \end{equation*}
    \begin{equation*} \text{or,}\quad 5T+0.321T = 64.2+150 \end{equation*}
    \begin{equation*} \therefore\quad T = \frac{214.2}{5.321} =40^{O}C =313 \,K \end{equation*}
  2. \begin{equation*} \,dS_{i} = \int\limits_{T_{i}}^{T}\frac{\,dQ}{T} = \int\limits_{T_{i}}^{T}m_{i}c_{i}\frac{\,dT}{T} = m_{i}c_{i} \ln\left(\frac{T}{T_{i}}\right) \end{equation*}
    \begin{equation*} \therefore\quad \,dS_{i} = (300 \,g) (0.107 \,cal/g^{o}C) \ln\left(\frac{40}{200}\right) \end{equation*}
    \begin{equation*} =51.4 \,cal/^{o}C = -55.4 \,J/K \end{equation*}
  3. \begin{equation*} \,dS_{w} = \int\limits_{T_{w}}^{T}\frac{\,dQ}{T} = \int\limits_{T_{w}}^{T}m_{w}c_{w}\frac{\,dT}{T} = m_{w}c_{w} \ln\left(\frac{T}{T_{w}}\right) \end{equation*}
    \begin{equation*} \therefore\quad \,dS_{w} = (500 \,g) (1 \,cal/g^{o}C) \ln\left(\frac{40}{30}\right) \end{equation*}
    \begin{equation*} =146.9 \,cal/^{o}C = 69.7 \,J/K \end{equation*}
  4. \begin{equation*} \,dS= \,dS_{i}+\,dS_{w} =-51.4+146.9=95.4 \,cal/^{o}C= 14 \,J/K \end{equation*}
  5. yes, entropy always increases in irreversible process.

Heat Transfer.

Example 10.4.10.

The wall of a house is 6 m high, 7 m wide and 0.3 m thick made out of brick of thermal conductivity, \(\kappa=0.6 \,W/mK.\) The interior temperature of the house is \(18^{o} C\) and that at the exterior is \(5^{o} C,\) find the rate of heat loss through the wall.
Solution.
Given: \(\kappa=0.6 \,W/mK,\) \(l=7 \,m, \quad b=6 \,m, \quad x=0.3 \,m,\) \(T_{h}=18^{o} C, \) and \(T_{c}=5^{o} C.\) Now the rate of heat transfers due to conduction,
\begin{equation*} H=\frac{A \kappa \left(T_{h}-T_{c}\right)}{x} \end{equation*}
\begin{equation*} = \frac{7\times 6\times 0.6\times(18-5)}{0.3} = 1092 \,W \end{equation*}

Example 10.4.11.

A carpenter builds the walls of a basement out of solid concrete 4 inches thick. In order to reduce the heat loss to the ground, he surrounds the wall with styrofoam sheets 2 inches thick before filling in the dirt. The average ground temperature at that location is \(40^{o}F\) and the interior temperature of the basement is kept at a comfortable \(75^{o}F\) year round.
  1. How much heat flows through the wall of the basement per square meter during the course of one year?
  2. If the basement measures 10 m long, 5 m wide, and 3 m high how much heat flows out the walls in one year?
  3. If electricity costs 10 cents per kilowatt-hour, how much would it cost to heat the basement per year? Note: \(k_{wood} = 0.8 \,W/m.K,\quad k_{styrofoam} = 0.01 \,W/m.K.\)
Solution.
Given: \(l_{1}= 4 \,in= 0.1016 \,m, \quad l_{2}= 2 \,in = 0.0508 \,m, \) \(T_{c}= 40^{o}F =4.44^{o}C,\quad T_{h}= 75^{o}F =23.89^{o} C,\) \(t = 1 \,y = 3.156\times 10^{7} \,s,\) \(k_{1} = 0.8 \,W/m K, \quad k_{2} = 0.01 \,W/m K.\)
  1. \(A_{1}= 1 \,m^{2},\)
    \begin{equation*} H_{1} = \frac{Q_{1}}{t} = \frac{\kappa_{1} A \left(T_{h}-T\right) }{l_{1}}, \end{equation*}
    \begin{equation*} H_{2} = \frac{Q_{2}}{t} =\frac{\kappa_{2} A \left(T-T_{c}\right) }{l_{2}} \end{equation*}
    Since rate of heat flow from the wall is constant,
    \begin{equation*} H_{1}=H_{2}=H. \end{equation*}
    We can find the value of T and plug that value in either \(H_{1}\) or \(H_{2}\) to get get \(Q\) by using formula
    \begin{equation*} Q=Ht \end{equation*}
    \begin{equation*} \text{Now,}\quad R = \left(\frac{l_{1}}{k_{1}} +\frac{l_{2}}{k_{2}} \right) \end{equation*}
    \begin{equation*} =\frac{A}{H}\left(T_{h}-T\right)+\frac{A}{H}\left(T-T_{c}\right) =\frac{A}{H}\left(T_{h}-T_{c}\right) \end{equation*}
    \begin{equation*} \therefore \quad H = \frac{Q}{t}= \frac{A (T_{h} -T_{c} )}{R} = 1.16\times 10^{8}\,J \end{equation*}
  2. \(a=10 \,m, \quad b=5 \,m,\quad h=3 \,m,\quad A=2(ah+bh)=2(a+b)h \) for 4 walls
    \begin{equation*} \therefore\quad Q' = QA=2Qh(a+b) \end{equation*}
    \begin{equation*} = 2(10+5)3\times 1.16\times 10^{8}=1.05\times10^{10} \,J =2.92\times10^{3}\,kW-h \end{equation*}
    \(\because\quad J=Ws=kwh/(1000\times 60\times60)\)
  3. \begin{equation*} c=10 \,cents/kW-h=\$0.1/kW-h \end{equation*}
    \begin{equation*} \therefore\quad C=Q'c= (2.92\times10^{3} \,kW-h ) \times (\$0.1/kW-h) = \$292.00 \end{equation*}

Example 10.4.12.

The wall of a house is 4 m wide and 3 m high. Except for the windows the wall consists of a sheet of rock \((R = 0.013 \,m^{2}.K/W)\text{,}\) \(3\frac{1}{2}\) inches of fiberglass insulation \((R = 2.53 \,m^{2}.K/W)\text{,}\) and wooden siding \((R = 0.22 \,m^{2}.K/W)\text{.}\) There are two windows in the wall, each 1 m wide and 1.5 m tall consisting of single pane glass \((R = 0.21 m^{2}.K/W)\text{.}\)
  1. If the temperature outside is \(-10^{o}C\) and inside is \(25^{o}C\text{,}\) how much heat flows out the wall and windows in 1 hour?
  2. If the conductivity of window glass is \(0.78 W/m.K\) and the window is 3.2 mm thick, what is the effective R-value of each of the two layers of "dead" air in contact with the window?
Solution.
Given: \(a_{1}=4 \,m, \quad b_{1}=3 \,m, \) \(R_{1}=0.013 \,m^{2}.K/W, \quad R_{2}=2.53 \,m^{2}.K/W, \) \(R_{3}=0.22 \,m^{2}.K/W,\) \(a_{2}=1 \,m, \quad b_{2}=1.5 \,m,\) \(R_{4}=0.21 \,m^{2}.K/W,\) \(T_{1} =-10^{o}C, \quad T_{2} =25^{o}C, \quad t=1 \,h=3600 \,s.\)
  1. \(A_{1}=a_{1}b_{1};\) \(R_{5}=R_{1}+R_{2}+R_{3}\) \(A_{2}=2a_{2}b_{2}\)
    \begin{equation*} H_{1}= \frac{(A_{1}-A_{2})\Delta T}{R_{5}}; \end{equation*}
    \begin{equation*} H_{2}= \frac{A_{2}\Delta T}{R_{4}} \end{equation*}
    \begin{equation*} H=H_{1} +H_{2};\quad H=\frac{Q}{t} \end{equation*}
    \(\Delta T = 35 \,k, \) \(A_{1}=12 \,m^{2},\) \(R_{5} =2.763 \,m^{2}K/W,\) and \(A_{2}=3 \,m^{2},\)
    \begin{equation*} H_{1}=114 \,W, \end{equation*}
    \begin{equation*} H_{2}=22 \,W, \end{equation*}
    \begin{equation*} H=136 \,W \end{equation*}
    \begin{equation*} \therefore\quad Q=4.898\times10^{5} \,J \end{equation*}
  2. \(k= 0.78 \,W/m.K, \) \(l= 0.0032 \,m, \)
    \begin{equation*} R=l/k \end{equation*}
    \begin{equation*} R_{4}=R+2R_{air}; \end{equation*}
    \begin{equation*} R=4.1\times10^{-3} \,m^{2}K/W \end{equation*}
    \begin{equation*} \therefore\quad R_{air}=0.103 \,m^{2}K/W \end{equation*}

Example 10.4.13.

Assume that the total surface area of the human body is about \(1 \,m^{2}\) and the surface temperature of the skin is \(30^{o}C\text{.}\)
  1. What is the rate of heat loss through radiation by the body if the emissivity of the body is 0.9?
  2. What is the net rate of heat loss through radiation by the body into an environment at \(20^{o}C\text{?}\)
Solution.
Given: \(A=1.2 \,m^{2}, \quad T_{1}=30^{o}C=303 \,K,\) \(e=0.9, \quad \sigma = 5.67\times10^{-8} \,W/m^{2}K^{4},\) \(T_{2}=15^{o}C= 288 \,K.\)
  1. \begin{equation*} P_{1}=\sigma A e T_{1}^{4} = 516.15 \,W \end{equation*}
  2. \begin{equation*} P_{2}=\sigma A e T_{2}^{4} =412.29 \,W \end{equation*}
    \begin{equation*} \therefore \quad P=P_{1}-P_{2} = 94.9 \,W \end{equation*}

Example 10.4.14.

Two metal rods, one lead and the other copper, are connected in end-to-end in series. Each rod is 0.5 m in length and has a square cross section 1.5 cm on a side. The temperature at the lead end of the rods is \(2^{o}C\text{;}\) the temperature at the copper end is \(106^{o}C\text{.}\)
  1. The average temperature of the two ends is \(54^{o}C\text{.}\) Is the temperature in the middle, at the lead-copper interface, greater than, less than, or equal to \(54^{o}C\text{?}\) Explain.
  2. Find the temperature at the lead-copper interface.
Given: \(\kappa_{Pb}=34.3 \,W/mK,\) and \(\kappa_{Cu}=395 \,W/mK.\)
Solution.
  1. Since \(\kappa_{Cu} \gt \kappa_{Pb}\) and the rate of heat flow through both of the rods must be equal, then temperature at the junction must be greater than the average temperature \(54^{o}C\text{,}\) otherwise lead can not dissipate the same amount of heat every second as the copper does.
    \begin{equation*} H_{Pb} =H_{Cu} \end{equation*}
    \begin{equation*} \text{or,}\quad \frac{\kappa_{Pb}A(\Delta T)_{Pb}}{l_{Pb}} = \frac{\kappa_{Cu}A(\Delta T)_{Cu}}{l_{Cu}} \end{equation*}
    \begin{equation*} \text{or,}\quad \kappa_{Pb}(\Delta T)_{Pb} = \kappa_{Cu}(\Delta T)_{Cu} \end{equation*}
    \begin{equation*} \text{or,}\quad \frac{\kappa_{Cu}}{\kappa_{Pb}} = \frac{(\Delta T)_{Pb}}{(\Delta T)_{Cu}} \end{equation*}
    \(\because\quad \kappa_{Cu} \gt \kappa_{Pb},\)
    \begin{equation*} \therefore\quad (\Delta T)_{Pb} \gt (\Delta T)_{Cu} \end{equation*}
    Hence, the lead requires a larger temperature difference across it than the copper, to get the same heat flow.
  2. \begin{equation*} H_{Pb} = H_{Cu} \end{equation*}
    \begin{equation*} \text{or,}\quad \kappa_{Pb}(\Delta T)_{Pb} = \kappa_{Cu}(\Delta T)_{Cu} \end{equation*}
    \begin{equation*} \text{or,}\quad \kappa_{Pb}(T-T_{o}) = \kappa_{Cu}(T_{h}-T) \end{equation*}
    \begin{equation*} \therefore\quad T = \frac{T_{h}\kappa_{Cu}+T_{o}\kappa_{Pb}}{\kappa_{Cu}+\kappa_{Pb}} = 98^{o}C \end{equation*}

Example 10.4.15.

A copper pipe of radius 10mm is carrying a hot water. The external surface of the pipe is subjected to a material of convective heat transfer coefficient of \(h=6W/m^{2}K,\) find the heat lost by the convection per unit length of the pipe when external surface temperature is \(80^{o}C\) and the surroundings are at \(20^{o}C.\) Assuming a black body radiation what is the heat lost by radiation?
Solution.
For 1 m long pipe,
\begin{equation*} H_{conv}=hA(T_{s}-T_{f}) \end{equation*}
\begin{equation*} = 6 (2\pi r)\times 1m \times (80-20) = 12(W/m^{2}K)\pi(10\times10^{-3}m)(60 K) = 22.6 W \end{equation*}
Now for radiation by assuming a pipe as black body
\begin{equation*} P=\sigma A \epsilon \left(T_{s}^{4}-T_{f}^{4}\right) \end{equation*}
\begin{equation*} =\sigma A \left(T_{s}^{4}-T_{f}^{4}\right) \end{equation*}
\begin{equation*} = (5.67\times10^{-8} \,W/m^{2}.K^{4})(2\pi 0.01\,m)(1\,m)\times(353^{4}-293^{4})\,K^{4} \end{equation*}
\begin{equation*} \therefore\quad H = P= 29.1 \,W \end{equation*}

Example 10.4.16.

A solid aluminum sphere coated with lampblack (emissivity=0.97) is suspended in an evacuated container. The sphere has a radius of 0.020 m and is initially at \(20^{o}C.\) The container is maintained at a temperature of \(70^{o}C.\)
  1. Assuming that the temperature of the sphere does not change very much, what is the net energy gained by the sphere in 10.0 s?
  2. Estimate the change in temperature of the sphere.
Solution.
  1. The energy lost must be supplied at the same rate at which it is being lost by radiation. At equilibrium, power emitted = power absorbed. Therefore, the energy lost per second by the sphere is
    \begin{equation*} \frac{\Delta Q}{\Delta t}=\left(P_{ab} -P_{em}\right) \end{equation*}
    \begin{equation*} \Delta Q = \sigma A \epsilon\left(T_{h}^{4}-T_{c}^{4}\right)\Delta t \end{equation*}
    \begin{equation*} \therefore\quad \Delta Q =(5.67\times10^{-8} \,W/m^{2}.K^{4})\times (4\pi r^{2})\,m^{2} \times(0.97) \end{equation*}
    \begin{equation*} \qquad \times(343^{4}-293^{4}\,K^{4})(10\,s) =18\,J \end{equation*}
    \([\because r=0.02 \,m]\)
  2. \begin{equation*} \Delta Q=mc\Delta T \end{equation*}
    \begin{equation*} \therefore\quad \Delta T = \frac{\Delta E}{mc} \end{equation*}
    \begin{equation*} =\frac{\Delta E}{V_{al}\rho_{al}c_{al}} \end{equation*}
    \begin{equation*} = \frac{18 \,J}{\left(\frac{4}{3}\pi\times 0.02^{3} \,m^{3}\right)(2700 \,kg/m^{3})(900 \,J/kgK)} = 0.22^{o}C. \end{equation*}