## College Physics (4th Edition)

The brass plate should be heated to a temperature of $26.75^{\circ}C$
We can find the required change in area of the hole: $\Delta A = 4.91000~cm^2-4.90874~cm^2 = 0.00126~cm^2$ We can find the required change in temperature of the brass plate so that the increase in area of the hole $\Delta A$ is $0.00126~cm^2$: $\Delta A = 2~\alpha~\Delta T~A$ $\Delta T = \frac{\Delta A}{2~\alpha~A}$ $\Delta T = \frac{0.00126~cm^2}{(2)(19~\times 10^{-6}~K^{-1})(4.90874~cm^2)}$ $\Delta T = 6.75~K$ $\Delta T = 6.75^{\circ}C$ The required increase in temperature is $6.75^{\circ}C$, so the brass plate should be heated to a temperature of $26.75^{\circ}C$