College Physics (4th Edition)

The copper washer should be heated to a temperature of $138^{\circ}C$
We can find the required change in diameter of the hole: $\Delta L = 1.0000~cm-0.9980~cm = 0.002~cm$ We can find the required change in temperature of the copper washer so that the increase in diameter of the hole $\Delta L$ is $0.002~cm$: $\Delta L = \alpha~\Delta T~L$ $\Delta T = \frac{\Delta L}{\alpha~L}$ $\Delta T = \frac{0.002~cm}{(17~\times 10^{-6}~K^{-1})(0.9980~cm)}$ $\Delta T = 118~K$ $\Delta T = 118^{\circ}C$ The required increase in temperature is $118^{\circ}C$, so the copper washer should be heated to a temperature of $138^{\circ}C$