Question

Liquid ammonia (boiling boint = -34.4 C) cn be used as a refrigerant and heat transfer fluid. How much energy is needed to heat 25.0 g of NH3(l) from -100 C to -10 C?

specific heat, NH3(l): 4.7 J/g K

specific heat, NH3(g): 2.2 J/g K

heat of vaporization: 23.5 kJ/mol

Answer 1

Heat needed to get from -100 to the boiling point -34.4= mass x specific heat of liquid x change of temp
= 25 x 4.7 x 65.5 Joules = 95.3 Joules ..............Part A

Heat needed to vaporize 25 g = mass x heat of vaporization = 25 g x (1 mol/17 g ammonia) x (23500 Joules/mol) = 34558.8    ........................Part B

Heat needed to get gaseous ammonia from -34.4 to -10 = mass x specific heat of gas x change of temp
= 25 x 2.2 x 24.4 Joules = 51.6   .................. Part C

The total amount of heat will be the sum of the heat in part a + part b + part c. = 95.3 + 34558.8+ 51.6 = 34705.7 Joules