Liquid ammonia can be used as a refrigerant and heat transfer fluid.How much energy is required to heat 25.0 g of NH₃(l) from -65.0°C to -12.0°C? Normal boiling point -33.4°C
Specific heat of NH₃(l) 4.7 J/g • °C
Specific heat of NH₃(g) 2.2 J/g • °C
Molar heat of vaporization 23.5 kJ/mol

Question

Quizplus · Accepted Answer

To calculate the total energy required, we need to consider three steps: heating the liquid ammonia from -65.0°C to its boiling point at -33.4°C, vaporizing it at the boiling point, and then heating the vapor from -33.4°C to -12.0°C. First, calculate the energy to heat the liquid: q1 = mass × specific heat (liquid) × temperature change = 25.0 g × 4.7 J/g°C × (31.6°C) = 3715 J. Next, calculate the energy for vaporization: q2 = moles × molar heat of vaporization = (25.0 g / 17.03 g/mol) × 23.5 kJ/mol = 34.5 kJ. Finally, calculate the energy to heat the vapor: q3 = mass × specific heat (gas) × temperature change = 25.0 g × 2.2 J/g°C × (21.4°C) = 1178 J. Convert all energies to kJ and sum them: (3.715 kJ + 34.5 kJ + 1.178 kJ) = 39.4 kJ.

Liquid Ammonia Can Be Used as a Refrigerant and Heat