Question

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?

Answer 1

Answer:

39.4 kJ

Explanation:

Given mass = 25.0 g

Specific heat of ammonia (g) = 4.7 J/g.K

Specific heat of ammonia (l) = 2.2 J /g.K

ΔHvap  = 23.5 x 10³ J /mol

Heating liquid ammonia from -65 °C to -33.4°C

Q₁ = m s ΔT  =  = 25.0 g * 4.7 J/g.K * [-33.4 - (-65)] °C = 3713 J

Evaporating 25 g of ammonia at -33.4°C

Molar mass of ammonia = 17 g/mol

Thus, moles = 25 / 17 moles

Q₂ = (25 / 17) mol * ΔHvap   = 1.470 mol * 23.5 x 10³ J /mol  = 34558.8 J

Heating NH₃(g) from -33.4°C  to -12.0°C

Q₃ = m s ΔT  =  = 25.0 g * 2.2 J/g.K * [-12.0 - (-33.4)] °C = 1177 J    

Total heat energy = Q₁ + Q₂ +Q₃  = 39448.8 J  = 39.4 kJ (As 1 J = 10⁻³ kJ)