College Physics (4th Edition)

$502,320~J$ of heat must flow into the water.
We can find the mass of the water: $m = (1000~kg/m^3)(2.0\times 10^{-3}~m^3) = 2.0~kg$ We can find the required heat: $Q = m~c~\Delta T$ $Q = (2.0~kg)(4186~J/kg~C^{\circ})(60.0~C^{\circ})$ $Q = 502,320~J$ $502,320~J$ of heat must flow into the water.