Thermal Properties of Matter - Result Question 25
26. 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, $\frac{d Q}{d t}$ through the rod in a steady state is given by:
[2009]
(a) $\frac{dQ}{dt}=\frac{k(T_1-T_2)}{LA}$
(b) $\frac{dQ}{dt}=kLA(T_1-T_2)$
(c) $\frac{dQ}{dt}=\frac{kA(T_1-T_2)}{L}$
(d) $\frac{dQ}{dt}=\frac{kL(T_1-T_2)}{A}$
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Answer:
Correct Answer: 26. (c)
Solution:
(c) $\frac{d Q}{d t}=\frac{k A(T_1-T_2)}{L}$
[ $(T_1-T_2)$ is the temperature difference $]$
Rate of heat flow is called heat current,
$H=\frac{d Q}{d t}$ and $H=\frac{(T_1-T_2)}{R}$
and thermal resistance, $R=\frac{L}{k A}$
