Question

The two ends of a rod of length $L$ and a uniform cross-sectional area $A$ are kept at two temperatures $T_1$ and $T_2 (T_1 > T_2).$ The rate of heat transfer, through the rod in a steady state is given by :

• A. $\frac{ dQ}{dt}= \frac{ k(T_1 -T_2)}{LA}$
• B. $\frac{ dQ}{dt}= k LA(T_1 -T_2)$
• C. $\frac{ dQ}{dt}= \frac{ kA(T_1 -T_2)}{L}$
• D. $\frac{ dQ}{dt}= \frac{ kL(T_1 -T_2)}{A}$

Answer 1

Answer: (C)

$\frac{ dQ}{dt}= \frac{ kA(T_1 -T_2)}{L}$

Answer 2

For steady state $\frac{ dQ }{ dt }=\frac{ kA \left( T _{1}- T _{2}\right)}{ L }$