Question

How much energy is released or absorbed when 1 gm of steam at 100°C turns to ice at 0°C? 

Answer 1

In CGS system : 

Mass of water = m = 1 gm 

Latent heat of vapourisation = L_vapour = 540 cal/gm 

Latent heat of fusion of ice = L_fusion = 80 cal 

Specific heat of water = S = 1 cal/gm-0°C 

Difference in temperature 

∆T = 100 – 0 = 100°C. 

The energy transferred when 1 gram of steam at 100°C turns to ice at 0°C

Q = mL_vapour  mS∆T + mL_fusion

= 1 × 540 + 1 × 1 × 100 + 1 × 80 = 540 + 100 + 80 = 720 cal. 

(OR) 

Mass of water = m = 1 gm = 1/1000 kg

Latent heat of vapourisation = L_vapour  2.25 × 10⁶ J/kg 

Latent heat of fusion = L_fusion= 3.36 × 10⁵ J/kg 

Difference in temperature 

∆T = 373 – 273 – 100 K. 

Specific heat of water = 4180 J/kg-K 

The energy transferred when 1 gram of steam at 100° C turns to ice at 0°C = Q = mL_vapour  mS∆T + mL_fusion

(OR)

In CGS system : 

Conversion : Steam at 100°C → Water at 100°C → 

Water at 0°C → Ice at 0°C. 

Mass of water = m = 1 gm 

Latent heat of vapourisation = L_vapour = 540 cal/gm 

Latent heat of fusion of ice = L_fusion= 80 cal 

Specific heat of water = S = 1 cal/gm-0°C 

Difference in temperature 

∆T= 100-0 = 100°C. 

The energy transferred when 1 gram of steam at 100°C turns to ice at 0°C 

Q = mL_vapour + mS∆T + mL_fusion 

= 1 × 540 + 1 × 1 × 100 + 1 × 80 

= 540 + 100 + 80 = 720 cal. 

= 720 × 4.18 = 3009.6 J.