Essential University Physics: Volume 1 (3rd Edition)

We find the total heat that is consumed: $Q = mL + mc\Delta T \\ Q=(.12)(334,000)+(.12)(4184)(37) \\ Q= 59 \times 10^3 J$ Thus, we find: $\Delta t = \frac{ 59 \times 10^3 J}{800W}=73s$