Question

How much heat is evolved in converting 1.00 mol of steam at 140.0 ∘C to ice at -45.0 ∘C? The heat capacity of steam is 2.01 J/(g⋅∘C) and of ice is 2.09 J/(g⋅∘C).

Answer 1

Sol :-

1 mole of steam = 18.00 g H₂O
latent heat of fusion of water which is = 336 kJ/kg .

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

Specific heat of water which is = 4.184 kJ/kg degree C .

Follow these steps :-

Step.1 :- Cooling the Steam from 140°C to 100°C.

Q = mass (m) x specific heat of steam (C_p) x ΔT (temperature difference)
Q = 0.018 kg x 2.01 kJ/kg/°C x 40°C = -1.4472 kJ (released).

Step.2 :- Condensing the steam at 100°C to water at 100°C. the formula is

Q = mass (m) x latent heat of vaporization of water
Q = 0.018 kg x 2,260 kJ/kg = - 40.68 kJ (released).

Step.3 :- Cooling the water from 100°C to 0°C.

Q = mass (m) x  specific heat of wate(C_p) x ΔT
Q = 0.018 kg x 4.184 kJ/kg/° C x 100°C = - 7.5 kJ (released).

Step.4 :- Using latent heat of fusion to turn water at 0 °C to ice at 0 °C (that is freezing water at 0 °C) .

Q = mass x  latent heat of fusion of water
Q = 0.018 kg x 336 kJ/kg = - 6.048 kJ (released)..

Step.5 :- Cooling the ice from 0 ^o C to - 45°C

Q = mass x specific heat of ice x ΔT
Q = 0.018 kg x 2.09 kJ/kg x 45°C = - 1.6929 kJ  (released).

Step.6  :-Total heat released  = -1.4472 kJ - 40.68 kJ - 7.5 kJ - 6.048 kJ - 1.6929 kJ = - 57.368 kJ (released)

Hence total amount of that can be evolved = 57.368 kJ