## College Physics (4th Edition)

The heat required to make the ice melt is equal in magnitude to the loss of heat from the aluminum as it cools: $m_a~c_a~\Delta T = m_i~L$ $m_a = \frac{m_i~L}{c_a~\Delta T}$ $m_a = \frac{(0.010~kg)(333\times 10^3~J/kg)}{(900~J/kg~C^{\circ})(80.0~C^{\circ})}$ $m_a = 0.04625~kg$ $m_a = 46.25~grams$ 46.25 grams of aluminum must be dropped in the hole.