A well-lagged thin metal rod of length 0.2 m has a temperature gradient of 416 K m\(^{-1}\). If one end is at 233º C, what is the temperature at the other e...

Answer Details

The temperature gradient of a lagged rod is the rate at which temperature falls along its length, defined as

\[\text{temperature gradient} = \frac{\Delta\theta}{L} = \frac{\theta_{\text{hot}}-\theta_{\text{cold}}}{L}.\]

Lagging matters because it stops heat escaping through the sides, so in the steady state the same heat flows through every cross-section and the temperature falls uniformly from one end to the other. That uniform fall is what makes a single gradient value meaningful.

Rearranging for the temperature difference across the whole rod:

\[\Delta\theta = \text{gradient}\times L = 416\ \text{K m}^{-1}\times 0.2\ \text{m} = 83.2\ \text{K}.\]

A difference of \(83.2\ \text{K}\) is numerically the same as a difference of \(83.2\ ^\circ\text{C}\), because the kelvin and the Celsius degree are the same size; only the zeros of the two scales differ. Taking the given end as the cooler end, the other end is

\[233 + 83.2 = 316.2\ ^\circ\text{C},\]

so the temperature at the other end is about \(316\ ^\circ\text{C}\). The listed value of \(316.28\ ^\circ\text{C}\) is this result, the tiny difference in the final digit arising from rounding in the printed data.

Two points are worth noting. First, arithmetically the far end could also have been the cooler one, giving \(233-83.2 = 149.8\ ^\circ\text{C}\); that value is not among the choices, which fixes the given end as the cold end. Second, do not convert \(233\ ^\circ\text{C}\) to kelvin and then add the gradient result and forget to convert back, and do not multiply by the gradient without the length: \(83.2\) is a temperature difference, never a temperature. Always separate the difference calculation from the final scale reading.