Two ends of a rod of length L and a uniform cross-sectional area A are kept at two temperatures T _{

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10-2.Transmission of Heat

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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 $T _{2}\left( T _{1} > T _{2}\right) .$ The rate of heat transfer, $\frac{ dQ }{ dt }$ through the rod in a steady state is given by

A

$\frac{ dQ }{ dt }=\frac{ k \left( T _{1}- T _{2}\right)}{ LA }$

B

$\frac{ dQ }{ dt }={kLA}\left( T _{1}- T _{2}\right)$

C

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

D

$\frac{ dQ }{ dt }=\frac{ kL \left( T _{1}- T _{2}\right)}{ A }$

Solution

The rate of heat transfer through the rod in a steady state is,

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

Standard 11

Physics

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