Question

Liquid nitrogen is often used as an effective coolant, as its boiling temperature is far below the freezing temperature of water. Specifically, the boiling point of liquid nitrogen is 77.0 K. A 0.650-kg block of iron at an initial temperature of 293.15 K is immersed in an insulated bath of liquid nitrogen with an initial temperature of 77.0 K. After the iron and the liquid nitrogen reach a state of thermal equilibrium, the iron block has cooled to a final a temperature of 77.0 K, and 203 g of liquid nitrogen has boiled off (vaporized) Assuming that the specific heat of iron over this temperature range is is 285 J/kg.K, find the latent heat of vaporization of liquid nitrogen

Answer 1

Liquid nitrogen has remained at 77 K initially and finally

Iron is cooled from 293.15 K to 77 K

hence heat lost by Iron is equal to the heat utilised for evaporation of Liquid nitrogen

specific heat s = Q/dM*dT = 285 J/kg-K for iron

heat lost by Iron = 285*(293.15-77)*0.65 = 40041.79 J

203 gm of liquid nitrogen evaporated at 77K

latent heat of evaporation = 40041.79/203 = 197.25 J/gm