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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, $\frac{dQ}{dt}$, through the rod in a steady state is given by
A
$\frac{dQ}{dt} = \frac{kA(T_1 - T_2)}{L}$
B
$\frac{dQ}{dt} = \frac{kL(T_1 - T_2)}{A}$
C
$\frac{dQ}{dt} = \frac{k(T_1 - T_2)}{LA}$
D
$\frac{dQ}{dt} = kLA(T_1 - T_2)$
Detailed Solution
In the steady state the rate of heat flow through a rod is proportional to the area of cross-section A and to the temperature difference $(T_1 - T_2)$, and inversely proportional to the length L.
$\frac{dQ}{dt} \propto \frac{A(T_1 - T_2)}{L}$
The constant of proportionality is the thermal conductivity k of the material.
$\frac{dQ}{dt} = \frac{kA(T_1 - T_2)}{L}$
$\frac{dQ}{dt} \propto \frac{A(T_1 - T_2)}{L}$
The constant of proportionality is the thermal conductivity k of the material.
$\frac{dQ}{dt} = \frac{kA(T_1 - T_2)}{L}$