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 $${T_2}\left( {{T_1} > {T_2}} \right).$$ The rate of heat transfer, $$\frac{{dQ}}{{dt}},$$ through the rod in a steady state is given by
A.
$$\frac{{dQ}}{{dt}} = \frac{{KL\left( {{T_1} - {T_2}} \right)}}{A}$$
B.
$$\frac{{dQ}}{{dt}} = \frac{{K\left( {{T_1} - {T_2}} \right)}}{{LA}}$$
C.
$$\frac{{dQ}}{{dt}} = KLA\left( {{T_1} - {T_2}} \right)$$
D.
$$\frac{{dQ}}{{dt}} = \frac{{KA\left( {{T_1} - {T_2}} \right)}}{L}$$
Answer :
$$\frac{{dQ}}{{dt}} = \frac{{KA\left( {{T_1} - {T_2}} \right)}}{L}$$
Solution :
For a rod of length $$L$$ and area of cross-section $$A$$ whose faces are maintained at temperatures $${T_1}$$ and $${T_2}$$ respectively.
Then in steady state the rate of heat flowing from one face to the other face in time $$t$$ is given by $$\frac{{dQ}}{{dt}} = \frac{{KA\left( {{T_1} - {T_2}} \right)}}{L}$$