The two ends of a metal rod are maintained at temperatures $100 ^\circ ext{C}$ and $110 ^\circ ext{C}$. The

Question

The two ends of a metal rod are maintained at temperatures $100 ^\circ	ext{C}$ and $110 ^\circ	ext{C}$. The rate of heat flow in the rod is found to be $4.0 	ext{ J/s}$. If the ends are maintained at temperatures $200 ^\circ	ext{C}$ and $210 ^\circ	ext{C}$, the rate of heat flow will be:

Accepted Answer

The rate of heat flow through a rod by conduction is given by $H = \frac{\Delta Q}{\Delta t} = \frac{KA\Delta T}{L}$, where $K$ is the thermal conductivity, $A$ is the cross-sectional area, $L$ is the length, and $\Delta T$ is the temperature difference between the ends.
In the first case, $\Delta T_1 = 110 ^\circ	ext{C} - 100 ^\circ	ext{C} = 10 ^\circ	ext{C}$. The rate of heat flow is $H_1 = 4.0 	ext{ J/s}$.
In the second case, $\Delta T_2 = 210 ^\circ	ext{C} - 200 ^\circ	ext{C} = 10 ^\circ	ext{C}$.
Since the temperature difference $\Delta T$ is the same in both cases and the dimensions and material of the rod remain unchanged, the rate of heat flow will be the same (assuming the thermal conductivity $K$ is essentially constant over this temperature range).
Therefore, $H_2 = H_1 = 4.0 	ext{ J/s}$.

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