First slide

Conduction

Question

The two ends of a rod of length L and a uniform cross-sectional area A are kept at two temperatures T₁ and T₂ (T₁ > T₂ ). The rate of heat transfer, $\frac{dQ}{dt}$ through the rod in a steady state is given by:

Moderate

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

$\frac{dQ}{dt} = kLA (T_{1} - T_{2})$

Solution

$\frac{Q}{t} = \frac{k A (T_{1} - T_{2})}{L}$

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