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)
