## College Physics (4th Edition)

$157~grams~$ of ice must melt to lower the temperature of the water to $0.0^{\circ}C$
We can find the mass of the water: $m_w = (1000~kg/m^3)(500~mL)(\frac{1~m^3}{10^6~mL}) = 0.50~kg$ The heat required to make the mass of ice melt is equal to the loss of heat from the water as it cools: $m_i~L = m_w~c~\Delta T$ $m_i = \frac{m_w~c~\Delta T}{L}$ $m_i = \frac{(0.50~kg)(4186~J/kg~C^{\circ})(25~C^{\circ})}{333\times 10^3~J/kg}$ $m_i = 0.157~kg$ $157~grams~$ of ice must melt to lower the temperature of the water to $0.0^{\circ}C$.