Question

How much heat is evolved in converting 1.00 mol of steam at 130.0 ∘C to ice at -50.0 ∘C? The heat capacity of steam is 2.01 J/(g⋅∘C) and of ice is 2.09 J/(g⋅∘C).

Answer 1

1 mole of steam = 18.00g H2O

There are six steps in this problem
1] cooling steam from 130.0 degree C to 100 degree C
2] latent heat [heat of vaporisation] in turning steam at 100 degree C to water at 100 degree C
3] cooling water at 100 degree C to 0 degree C
4] using latent heat of fusion to turn water at 0 degree C to ice at 0 degree C
5] cooling ice from 0 degree C to -50 degree C

6] total heat evolved

and for solving the problem we need some extra values like latent heat of fusion of water which is = 336 kJ/kg .

latent heat of vaporization of water which is = 2260 kJ/kg .

and the specific heat of water which is = 4.184 kJ/kg degree C .

now the six steps are as follows.

1...Cooling the Steam from 130°C to 100°C. the formula is Q = mass * specific heat of steam * delta T (temperature difference)
= 0.018kg x 2.01kJ/kg/°C x 30°C = 1.0854kJ. Released.

2...Condensing the steam at 100°C to water at 100°C. the formula is Q = mass * latent heat of vaporization of water
= 0.018kg x 2,260kJ/kg = 40.7kJ. Released.

3...Cooling the water from 100°C to 0°C. and the formula is Q = mass * specific heat of water * delta T
= 0.018kg x 4.184kJ/kg/° C x 100°C = 7.5kJ. Released.

4..Using latent heat of fusion to turn water at 0 degree C to ice at 0 degree C (that is freezing water at zero degree C) . and the formula is Q = mass * latent heat of fusion of water
= 0.018kg x 336kJ/kg = 6.048 kJ. Released.

5...Cooling the ice from 0 ^o C to -50°C. and the formula is Q = mass * specific heat of ice * delta T
= 0.018kg x 2.09kJ/kg x 50°C = 1.881 kJ. Released.

6...Total heat released: = 1.881 kJ + 6.048 kJ + 7.5kJ + 40.7kJ + 1.0854kJ = 57.2144 kJ