10 kg of liquid water are cooled from an initial temperature of 40°C. Up until 0°C, an energy balance is used to determine the final temperature \(T_2\): $$ m_L\;Cp\;dT = - \Delta H_{vap,20}\;dm_v $$ where \( Cp \) is the liquid heat capacity (kJ/kg \(^\circ \)C), \( \Delta H_{vap,20} \) is the heat of vaporization at 20°C (average of initial temperature and 0°C) in kJ/kg, and \( T_2 \) is in °C.
The amount of liquid \( m_L \) is 10 kg minus the amount of water vaporized \( m_v \) (both in kg): $$ (10 - m_v) Cp\; dT = - \Delta H_{vap,20}\; dm_v $$ Integrating between the initial conditions (\( m_L = 10, m_v = 0 \)) and the final conditions yields: $$ \int_{40}^{T_2} \frac{Cp dT}{\Delta H_{vap,20}} = -\int_{0}^{m_v} \frac{dm_v}{(10-m_v)} $$ $$ \frac{Cp}{\Delta H_{vap,20}} (T_2 - 40) = \ln (10 - m_v) \Big|_{0}^{m_v} = \ln \left( \frac{10-m_v}{10} \right) $$ Thus the final temperature for a given amount of water vaporized is: $$ T_2 = 40 + \frac{\Delta H_{vap,20}}{Cp} \ln \left( \frac{10-m_v}{10} \right), \text{ where } T_2 \geq 0^\circ C $$ When 0.66 kg of water vaporizes, the liquid temperature is 0°C and additional vaporization freezes some of the remaining liquid. An energy balance at 0°C determines the amount of liquid that forms solid ice: $$ m_s \Delta H_f = (m_v - 0.66) \Delta H_{vap,0} $$ where \( m_s \) is the mass of ice formed (kg), \( \Delta H_f \) is the heat of fusion (kJ/kg) and \( \Delta H_{vap,0} \) is the heat of vaporization at 0°C (kJ/kg).