1.6 Mechanisms of heat transfer  (Page 4/27)

Note that in [link] , most of the best thermal conductors—silver, copper, gold, and aluminum—are also the best electrical conductors, because they contain many free electrons that can transport thermal energy. (Diamond, an electrical insulator, conducts heat by atomic vibrations.) Cooking utensils are typically made from good conductors, but the handles of those used on the stove are made from good insulators (bad conductors).

Two conductors end to end

Strategy

We repeat the calculation with a second method, in which we use the thermal resistance R of the rod, since it simply adds when two rods are joined end to end. (We will use a similar method in the chapter on direct-current circuits.)

Solution

1. Identify the knowns and convert them to SI units.
   The length of each rod is $L_{\text{A1}}=L_{\text{steel}}=0.25\text{m,}$ the cross-sectional area of each rod is $A_{\text{A1}}=A_{\text{steel}}=7.85\times {10}^{\mathrm{-5}}{\text{m}}^2\text{,}$ the thermal conductivity of aluminum is $k_{\text{A1}}=220\text{W/m}·\text{°}\text{C}$ , the thermal conductivity of steel is $k_{\text{steel}}=80\text{W/m}·\text{°}\text{C}$ , the temperature at the hot end is $T=100\text{°}\text{C}$ , and the temperature at the cold end is $T=20\text{°}\text{C}$ .
2. Calculate the heat-conduction rate through the steel rod and the heat-conduction rate through the aluminum rod in terms of the unknown temperature T at the joint:
   
   $\begin{array}{cc}\hfill P_{\text{steel}}& =\frac{k_{\text{steel}}{A}_{\text{steel}}\text{Δ}{T}_{\text{steel}}}{L_{\text{steel}}}\hfill \\ & =\frac{\left(80\text{W/m}·\text{°}\text{C}\right)\left(7.85\times {10}^{\mathrm{-5}}{\text{m}}^2\right)\left(100\text{°}\text{C}-T\right)}{0.25\text{m}}\hfill \\ & =\left(0.0251\text{W/}\text{°}\text{C}\right)\left(100\text{°}\text{C}-T\right);\hfill \end{array}$
   
   
   $\begin{array}{cc}\hfill P_{\text{A1}}& =\frac{k_{\text{Al}}{A}_{\text{A1}}\text{Δ}{T}_{\text{Al}}}{L_{\text{A1}}}\hfill \\ & =\frac{\left(220\text{W/m}·\text{°}\text{C}\right)\left(7.85\times {10}^{\mathrm{-5}}{\text{m}}^2\right)\left(T-20\text{°}\text{C}\right)}{0.25\text{m}}\hfill \\ & =\left(0.0691\text{W/}\text{°}\text{C}\right)\left(T-20\text{°}\text{C}\right).\hfill \end{array}$
3. Set the two rates equal and solve for the unknown temperature:
   
   $\begin{array}{ccc}\hfill \left(0.0691\text{W/}\text{°}\text{C}\right)\left(T-20\text{°}\text{C}\right)& =\hfill & \left(0.0251\text{W/}\text{°}\text{C}\right)\left(100\text{°}\text{C}-T\right)\hfill \\ \hfill T& =\hfill & 41.3\text{°}\text{C}.\hfill \end{array}$
4. Calculate either rate:
   
   $P_{\text{steel}}=\left(0.0251\text{W/}\text{°}\text{C}\right)\left(100\text{°}\text{C}-41.3\text{°}\text{C}\right)=1.47\text{W}.$
5. If desired, check your answer by calculating the other rate.

Solution

1. Recall that $R=L\text{/}k$ . Now $P=A\text{Δ}T\text{/}R,\text{or}\text{Δ}T=PR\text{/}A.$
2. We know that $\text{Δ}{T}_{\text{steel}}+\text{Δ}{T}_{\text{Al}}=100\text{°}\text{C}-20\text{°}\text{C}=80\text{°}\text{C}$ . We also know that $P_{\text{steel}}=P_{\text{Al}},$ and we denote that rate of heat flow by P . Combine the equations:
   
   $\frac{P{R}_{\text{steel}}}{A}+\frac{P{R}_{\text{Al}}}{A}=80\text{°}\text{C}.$
   
   Thus, we can simply add R factors. Now, $P=\frac{80\text{°}\text{C}}{A\left(R_{\text{steel}}+R_{\text{Al}}\right)}$ .
3. Find the $R_s$ from the known quantities:
   
   $R_{{}_{\text{steel}}}=3.13\times {10}^{\mathrm{-3}}{\text{m}}^2·\text{°}\text{C/W}$
   
   and
   
   $R_{{}_{\text{Al}}}=1.14\times {10}^{\mathrm{-3}}{\text{m}}^2·\text{°}\text{C/W}.$
4. Substitute these values in to find $P=1.47\text{W}$ as before.
5. Determine $\text{Δ}T$ for the aluminum rod (or for the steel rod) and use it to find T at the joint.
   
   $\text{Δ}{T}_{\text{Al}}=\frac{P{R}_{\text{Al}}}{A}=\frac{\left(1.47\text{W}\right)\left(1.14\times {10}^{\mathrm{-3}}{\text{m}}^2·\text{°}\text{C/W}\right)}{7.85\times {10}^{\mathrm{-5}}{\text{m}}^2}=21.3\text{°}\text{C,}$
   
   so $T=20\text{°}\text{C}+21.3\text{°}\text{C}=41.3\text{°}\text{C}$ , as in Solution $1$ .
6. If desired, check by determining $\text{Δ}T$ for the other rod.