91Ó°ÊÓ

Outside end of the copper rod temperature$={100}^{∘}C$

Outside end of the iron rod temperature$=0^{∘}C$

In order to solve this problem, we must recognize that since the welding point is at thermal equilibrium, i.e. its temperature is constant, both the copper and iron rods transmit the same amount of power.

Knowing that the two rods have equal cross-section areas $A$and length $L$we can therefore write,

$P_c=P_i$

$\frac{k_{c}A}{L}\mathrm{Δ}{T}_c=\frac{k_{i}A}{L}\mathrm{Δ}{T}_i$

Simplifying the areas and the lengths, we are left with

$k_c\mathrm{Δ}{T}_c=k_i\mathrm{Δ}{T}_i$

We are given in the problem that the copper end is at the highest temperature, let it be denoted by $T_2=373\mathrm{K}$, and the iron end is at the lowest temperature, let it be $T_1=273\mathrm{K}$.

Therefore, letting the temperature of the weld be $T$, the temperature differences will be,

$\mathrm{Δ}{T}_c=T_2-T$

$\mathrm{Δ}{T}_i=T-T_i$

Substituting the expressions for the temperature differences, we have

$k_c\left(T_2-T\right)=k_i\left(T-T_1\right)$

From this we can express the weld temperature as

$T\left(k_c+k_i\right)=T_2{k}_c+T_1{k}_i⇒T=\frac{T_2{k}_c+T_1{k}_i}{k_c+k_i}$

Referring to table 19.5 on the book, our thermal conductivities are $k_c=400\frac{\mathrm{W}}{\mathrm{mK}}$and $k_i=80\frac{\mathrm{W}}{\mathrm{mK}}$

Substituting these and the temperatures, we can find the weld temperature to be,

$T=\frac{373·800+273·80}{400+80}=356\mathrm{K}=83°\mathrm{C}$

Hence, the temperature at the midpoint where the rods are joined together is${83}^{∘}C.$