Liquid ammonia (boiling point = −33.4°C) can be used as a refrigerant and heat transfer fluid. How much energy is needed to heat 25.0 g of NH₃(l) from −65.0°C to −12.0°C? A)5.5 kJ B)6.3 kJ C)39 kJ D)340 kJ E)590 kJ

Question

Quizplus · Accepted Answer

The first step is to calculate the heat required to raise the temperature of the ammonia from -65.0°C to its boiling point at atmospheric pressure (-33.4°C).

q1 = mcΔT 
q1 = (25.0 g)(4.702 J/g°C)(33.4°C - (-65.0°C)) 
q1 = 39,027 J = 39 kJ

The second step is to calculate the heat required to vaporize the ammonia at its boiling point:

q2 = nΔHvap 
q2 = (25.0 g)/(17.03 g/mol) (23.3 kJ/mol) 
q2 = 34,217 J = 34.2 kJ

Note that we used the molar mass of ammonia (17.03 g/mol) and the molar enthalpy of vaporization (23.3 kJ/mol).

The third step is to calculate the heat required to lower the temperature of the vaporized ammonia from -33.4°C to -12.0°C:

q3 = mcΔT 
q3 = (25.0 g)(1.448 J/g°C)(-12.0°C - (-33.4°C)) 
q3 = 4,457 J = 4.46 kJ

The total heat required is the sum of q1, q2, and q3:

qtotal = q1 + q2 + q3 
qtotal = 39 kJ + 34.2 kJ + 4.46 kJ 
qtotal = 77.6 kJ

Therefore, the answer is C) 39 kJ.

Liquid Ammonia (Boiling Point = −33