## College Physics (4th Edition)

The heat capacity of 30.0 kg of ice is $63,000~J$
The heat capacity is the amount of heat required to raise the temperature by $1~ C^{\circ}$. We can find the heat capacity of 30.0 kg of ice: $Q = m~c~\Delta T$ $Q = (30.0~kg)(2100~J/kg~C^{\circ})(1~C^{\circ})$ $Q = 63,000~J$ The heat capacity of 30.0 kg of ice is $63,000~J$