50.0 g of ice was initially at −4.0°C and 1 atm. At 1 atm, it absorbed some heat and became 20.0 g of ice and 30.0 g of liquid water in equilibrium with each other. How much heat was absorbed?

Let’s first consider the processes that occurred with the initial ice. Firstly, the ice was heated to 0°C (melting point of water). Then, the part of ice melted to form the water, still at 0°C. As not all the ice melted, we can assume that the final temperature of ice and water is still 0°C. In summary, what happened is: Heating of ice -4°C  0°C. Melting of the part of ice. Now, we can describe the processes with equations of thermochemistry: Q=cm∆T, where Q is the heat transferred to the ice, c is the specific heat of the ice (2.050 kJ /(kg·K), m is the mass if the ice (50 g, or 0.05 kg) and ∆T is the change in its temperature. Remember, difference in temperature in °C equals the difference in temperature in K. Q=2.050(kJ/(kg·K))·0.05(kg)·(0-(-4))(K)= 0.41 kJ For the phase change, the equation is different, as we often use the approximation that the temperature doesn’t change during the phase change. Q=mL_f, where m is the mass of ice that melts (30 g, or 0.03 kg), L_f is the heat of fusion (3.33·102 kJ/kg). Thus, heat transferred to the ice at the second process is: Q= 0.03 (kg)·3.33·10^2 (kJ/kg)=9.99 kJ Finally, the sum of two heats is: Q_Σ=0.41kJ+9.99 kJ=10.4 kJ Answer: 10.4 kJ