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 transf - Physics MCQ Physics | ExamTest.live

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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{d Q}{d t}, through the rod in a steady state is given by:

A $\frac{d Q}{d t} = \frac{k ( T _{1} - T _{2} )}{L A}$
B $\frac{d Q}{d t} = k L A (T_{1} - T_{2})$
C $\frac{d Q}{d t} = \frac{k A ( T _{1} - T _{2} )}{L}$
D $\frac{d Q}{d t} = \frac{k L ( T _{1} - T _{2} )}{A}$

Solution & Step-by-step Explanation

According to Fourier's law of heat conduction, the rate of heat flow \frac{d Q}{d t} through a material is proportional to the area A and the temperature gradient \frac{Δ T}{L} . \frac{d Q}{d t} = \frac{k A ( T _{1} - T _{2} )}{L} where k is the thermal conductivity of the material.

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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{d Q}{d t}, through the rod in a steady state is given by:

\frac{d Q}{d t} = \frac{k ( T _{1} - T _{2} )}{L A}

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

\frac{d Q}{d t} = \frac{k A ( T _{1} - T _{2} )}{L}

\frac{d Q}{d t} = \frac{k L ( T _{1} - T _{2} )}{A}

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Thermal Properties Heat Transfer Conduction

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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, $\frac{d Q}{d t}$ , through the rod in a steady state is given by:

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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 T1​ and T2​ (T1​>T2​). The rate of heat transfer, dtdQ​, through the rod in a steady state is given by:

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Discussion

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: