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PROBLEM: Liquid ammonia, NH3(l), was once used in home refrigerators as the heat transfer fluid. The specific heat capacity of the liquid is 4.7 J/g•K and that of the vapor is 2.2 J/g•K. The enthalpy of vaporization is 23.33 kJ/mol at the boiling point. If you heat 12 kg of liquid ammonia from –50°C to its boiling point of –33.3°C, allow it to evaporate, and then continue warming to 0.0°C, how much energy must you supply? Step 1: Figure out how much heat is needed to warm the liquid to its boiling point. qliquid = (12 kg x 1000 g/1 kg)(4.7 J/g•K)(–33.3°C – –50.0°C) = 941,880 J x 1 kJ/1000 J = 9.4188 x 102 kJ – > 9.4 x 102 kJ (sig. digits) Step 2: Figure out how much heat is needed to evaporate the liquid to its boiling point. qevaporation = (12 kg x 1000 g/1 kg)(1 mol/17.0 g)(23.33 kJ/mol) = 16468 kJ – > 1.6 x 104 kJ (sig. digits) Step 3: Figure out how much heat is needed to warm the gas. qevaporation = (12 kg x 1000 g/1 kg)(2.2 J/g•K)(0.0°C – –33.3°C) = 879120 J x 1 kJ/1000 J = 8.7912 x 103 kJ – > 8.8 x 102 kJ (sig. digits) Step 4: Add all the heats together qtotal = 9.4 x 102 kJ + 1.6 x 104 kJ + 8.8 x 102 kJ = 1.782 x 104 kJ = 1.8 x 104 kJ