Question

Liquid ammonia, NH3(l), was once used in home refrigerators as the heat transfer fluid. The specific heat of the liquid is 4.7 J/g.K and that of the vapor is 2.2 J/g.K. The enthalpy of vaporization is 23.33 kJ/mol at the boiling point. If you heat 12 kg of liquid ammonia from -50.0 °C to its boiling point of -33.3 °C, allow it to evaporate, and then continue warming to 0.0 °C, how much heat energy must you supply?

Answer 1

Lquid ammonia gains sensible heat from -50 deg.c to -33.3 deg.c (-33.3 deg.c is its boiling point)

1. Sensible heat ( from -50 deg.c to -33.3 deg.c)= mass* specific heat of liquid * temperature difference

= 12*4.7*(-33.3+50) joules=941.88 joules

2. Liquid at -33.3 deg.c gets converted to vapor at -33.3 deg.c when latent heat is supplied

Latent heat supplied = mass* Latent heat of vaporization

Latent heat of vaporization is given in moles and hence mass has to be converted into moles

Moles of liquid ammonia =12/17 ( 17 is the molecular weight of ammonia)=0.7059

Latent heat to be supplied= 0.7059*23.33*1000 j= 16468.65 joules

3. From -33.3 deg.c vapor, the vapor has to be heated to 0 deg,c by transferring sensible heat

Sensbile heat of vapir= 12*2.2*(0+33.3)= 879.12 joules

Total heat energy to be supplied = sum of 1+2+3= 941.88 +16468.65+879.12=18289.65 joules