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Question

(A) The rate of heat conduction through a rod is given by $H = \frac{K A \Delta T}{l}$,where $K$ is the thermal conductivity,$A = \pi r^2$ is the cros

Vedclass · Accepted Answer

$r = 2 \, cm, l = 0.5 \, m$

Vedclass · Answer

(A) The rate of heat conduction through a rod is given by $H = \frac{K A \Delta T}{l}$,where $K$ is the thermal conductivity,$A = \pi r^2$ is the cross-sectional area,and $l$ is the length.Since the material is the same,$K$ is constant. For a fixed temperature difference $\Delta T$,the rate of heat flow is proportional to $\frac{r^2}{l}$.Let $X = \frac{r^2}{l}$. We calculate $X$ for each option:$A: X = \frac{(2)^2}{0.5} = \frac{4}{0.5} = 8$$B: X = \frac{(2)^2}{2} = \frac{4}{2} = 2$$C: X = \frac{(0.5)^2}{0.5} = 0.5$$D: X = \frac{(1)^2}{1} = 1$Comparing the values,option $A$ has the highest value of $X$,meaning it will conduct the most heat.

Four rods of the same material with different radii $r$ and length $l$ are used to connect two reservoirs of heat at different temperatures. Which one will conduct the most heat?

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