## College Physics (4th Edition)

The required increase in temperature would be $587~K$
We can find the required change in temperature of the aluminum so that the increase in circumference is 26 cm: $\Delta L = \alpha~\Delta T~L$ $\Delta T = \frac{\Delta L}{\alpha~L}$ $\Delta T = \frac{0.26~m}{(25~\times 10^{-6}~K^{-1})(17.72~m)}$ $\Delta T = 587~K$ The required increase in temperature would be $587~K$.