Consider two rods of the same length and different specific heats $\left(S_1, S_2\right)$, conductivities $\left(K_1, K_2\right)$ and area of cross sections $\left(A_1, A_2\right)$ and both having temperature $T_1$ and $T_2$ at their ends. If the rate of loss of heat due to conduction is equal then:

Question

Aicharya · Accepted Answer

Rate of conduction of heat is given as
$\frac{\Delta \mathrm{Q}}{t}=\frac{K A}{L}\left(T_2-T_1\right)$
According to the question
$\begin{aligned}
& \frac{\Delta Q_1}{\Delta t}=\frac{\Delta Q_2}{\Delta t} \\
& \Rightarrow \frac{K_1 A_1}{L}\left(T_1-T_2\right)=\frac{K_1 A_1}{L}\left(T_1-T_2\right) \\
& \Rightarrow K_1 A_1=K_2 A_2
\end{aligned}$ .

Consider two rods of the same length and different specific heats $\left(S_1, S_2\right)$, conductivities…

Solution