First slide

Conduction

Question

The two ends of a metal rod are maintained at temperatures 100⁰C and 110⁰C. The rate of heat flow in the rod is found to be 4 ${Js}^{- 1}$ . If the ends are maintained at temperatures 200⁰C and 210⁰C, the rate of heat flow will be

Moderate

A

44 ${Js}^{- 1}$
B

16.8 ${Js}^{- 1}$
C

8 ${Js}^{- 1}$
D

4 ${Js}^{- 1}$

Solution

Here, $Δ T_{1} = 110 - 100 = 10^{\circ} C$
$\frac{d Q_{1}}{d t} = 4 {Js}^{- 1}$
As the rate of heat flow is directly proportional to the temperature difference and the temperature difference in
both the cases is same, i.e. $∆ T_{2}$ = 210 $-$ 200 = 10⁰C. So, the same rate of heat will flow in the second case.
Hence, $\frac{d Q_{2}}{d t} = 4 {Js}^{- 1}$

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