Question 26

One end of a cylindrical conducting rod is maintained at $$T_1$$ and the other end at $$T_2$$. The amount of heat transferred from hotter end to colder end in 2 minutes is $$200\text{ J}$$. If the radius and the length of the rod were doubled and same temperature difference were to be maintained between the ends, find the amount of heat (in J) transferred through this rod in 5 minutes. Given that heat transferred by a conducting rod is given by $$Q = \frac{KA(T_1 - T_2)t}{L}$$ where $$Q$$ is the amount of heat transferred in time $$t$$ by a conductor of length $$L$$ and cross section $$A$$ when its ends are maintained at $$T_1$$ and $$T_2$$. $$K$$ is the coefficient of thermal conductivity which is a property of the material of the conductor. (Ignore heat losses from the curved surfaces of the conductor)


Correct Answer: 1000

Solution

$$Q$$ is proportional to $$\frac{At}{L}$$. Doubling the radius makes $$A$$ four times, doubling $$L$$ halves $$Q$$, and the time changes from 2 to 5 minutes. So $$Q' = 200 \times 4 \times \frac{1}{2} \times \frac{5}{2} = 1000\text{ J}$$.

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