## Physics: Principles with Applications (7th Edition)

The brass rod must be heated to ${8.1\times10^{2}}^{\circ}C$ to be $1.5%$ longer than it was at $25^{\circ}C$.
Let $l=$ the length of the brass rod at $25^{\circ}C$. $$\Delta l=al_{0}\Delta T$$$$1.015l-l=(19\times10_{-6}\frac{1}{^{\circ}C})(l)(T_{f}-25^{\circ}C)$$$$.015=(19\times10_{-6}\frac{1}{^{\circ}C})(T_{f}-25^{\circ}C)$$ $$T_{f}=814.47^{\circ}C$$