Question

The thickness of each plate is L1 = L2 =3 x 10-3

and metal plate 1 is at 100oC while plate 2 at

0oC. The surface A of metal plate 1 has dimensions 4cm x 2cm. Given that the thermal

conductivity of metal plate 1 is 48.1W/m.K and that of plate 2 is 68.2W/m.K, what is the

temperature of the soldered interface if there is a steady flow of heat from metal plate 1 to

plate2?

41.4oC

Answer 1

Solution;

The given data;

$L_1=L_2=3×10^{-3}m$

$A_1=A_2=4cm×2cm$

$k_1=48.1\frac{W}{mK}$

$k_2=68.2\frac{W}{mK}$

$T_1=100°c$

$T_2=0°c$

Let T be the temperature of the interface.

Rate of energy flow is given as;

$\frac Qt=\frac{kAdT}{L}$

At steady state,the energy flow between the two metals is the same.

Hence equate energy flow rate for both metals as;

$\frac{k_1A_1(T_1-T)}{L_1}=\frac{k_2A_2(T-T_2)}{L_2}$

But,

$A_1=A_2$ and $L_1=L_2$

Hence we have;

$k_1(T_1-T)=k_2(T-T_2)$

Substituting the values,we have;

$48.1(373-T)=68.2(T-273)$

$17941.3-48.1T=68.2T-18618.6$

$116.3T=36559.9$

$T=314.359K$

Convert Kelvin into degrees;

$T=314.36-273$

$T=41.35°c$