## Conceptual Physics (12th Edition)

The plant dissipates $Q = 1.5\times 10^{8} J$ of heat to a reservoir of water each second, which raises its temperature by 3 Celsius degrees. Each second, $Q = cm\Delta T = (\frac{4184 J}{kg^{\circ} C}) (m)(3^{\circ}C)$. Solve for m = 11950 kg.