## Essential University Physics: Volume 1 (3rd Edition)

We first must find the mass: $m = \rho V = 10^3 \times 420 = 4.2 \times 10^5 \ kg$ We know that the change in temperature is 80 degrees Celsius, and we know that 10 percent of 3 GW is $300\times10^6\ W$. Thus, we find: $\Delta t = \frac{4.2\times10^5 \times 4184 \times 80}{300 \times 10^6} = 470 \ s$