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Thermal conduction
Appears in
Concepts tested here
- Conduction and latent heat
- Heat conducted through a rod
- Heat current depends on temperature difference
- Rate of heat flow
All Questions
2015 AIPMT-I 1 question
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The two ends of a metal rod are maintained at temperatures $100^\circ C$ and $110^\circ C$. The rate of heat flow in the rod is found to be 4.0 J/s. If the ends are maintained at temperatures $200^\circ C$ and $210^\circ C$, the rate of heat flow will beRate of heat conduction: $\frac{dQ}{dt} = \frac{KA(T_2 - T_1)}{L} \propto (T_2 - T_1)$
The temperature difference is $10^\circ C$ in both cases.
So the rate of heat flow remains 4.0 J/s.
2012 AIPMT-MAINS 1 question
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A slab of stone of area 0.36 $m^2$ and thickness 0.1 m is exposed on the lower surface to steam at $100^\circ C$. A block of ice at $0^\circ C$ rests on the upper surface of the slab. In one hour 4.8 kg of ice is melted. The thermal conductivity of slab is (Given latent heat of fusion of ice = $3.36\times10^5$ J $kg^{-1}$)Heat conducted through the slab in time t melts the ice: $Q = \frac{KA(\Delta T)t}{L} = mL_f$
$K = \frac{mL_f\times L}{A(\Delta T)t} = \frac{4.8\times3.36\times10^5\times0.1}{0.36\times100\times3600}$
$K = \frac{161280}{129600} = \frac{56}{45} = 1.24$ J/m/s/$^\circ C$
2010 AIPMT-PRE 1 question
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A cylindrical metallic rod in thermal contact with two reservoirs of heat at its two ends conducts an amount of heat Q in time t. The metallic rod is melted and the material is formed into a rod of half the radius of the original rod. What is the amount of heat conducted by the new rod, when placed in thermal contact with the two reservoirs in time t?Heat conducted in time t: $Q = \frac{KA(T_1 - T_2)t}{L}$, so for the same material, temperatures and time, $Q \propto \frac{A}{L}$
New radius = $\frac{r}{2}$, so new area $A' = \pi\left(\frac{r}{2}\right)^2 = \frac{A}{4}$
The volume of metal is unchanged: $A'L' = AL$, so $L' = \frac{AL}{A/4} = 4L$
$\frac{Q'}{Q} = \frac{A'}{A}\times\frac{L}{L'} = \frac{1}{4}\times\frac{1}{4} = \frac{1}{16}$
$Q' = \frac{Q}{16}$
2009 AIPMT 1 question
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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 byIn 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.
$\frac{dQ}{dt} = \frac{kA(T_1 - T_2)}{L}$