## Essential University Physics: Volume 1 (4th Edition)

We first find the value of Q: $Q = mc \Delta t = (150)(4184)(50-18) \approx 20 \ MJ$ We now find the change in time: $\Delta t = \frac{20 \times 10^6 \ J }{5 \times 10^3 \ J/S} = 4000 \ s$