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

We first must find the amount of energy supplied by the burner over this time. Recalling that a Watt is a Joule per second, it follows: $2000 \frac{J}{s} \times 5.4 min \times \frac{60s}{min} = 648,000\ J$ Now, we set this equal to the equation for $\Delta Q$: $648,000\ J = 1.2 (447)(80) + (4184)(80)m \\ m = 1.8 \ kg$