Answer
The heat capacity of 30.0 kg of ice is $63,000~J$
Work Step by Step
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 $63,000~J$
