Question

The increase in internal energy of 1 kg of water at 100⁰C when it is converted into steam at the same temperature and at 1 atm (100 k Pa) will be [The density of water and steam are 1000 kg/m3 & 0.6 kg/m3 respectively. The latent heat of vapourisation of water is 2.25 × 106 J/kg.]

• A. 2.08 × 106 J
• B. 4 × 107 J
• C. 3.27 × 108 J
• D. 5 × 109 J

Answer 1

Answer: (A)

2.08 × 106 J

Answer 2

Latent heat of vaporisation of water = 2.25 × 106 J/kg

$\Delta$H = 2.25 × 106 J/kg.

work done = – Pext (V2 – V1)

$\Delta$H = 2.25 × 106 J/kg

$\Delta$H = $\Delta$U + PDV

(1) Now, volume of water V = $\left( \frac{m}{d} \right)$ = $\frac{1}{1000}$ M3 = 1L

(2) volume of steam = $\frac{1000}{0.6}$= 1666.67 L

2.25 × 106 = $\Delta$U + 1 [1666.67 – 1] 101.325

$\Delta$U = 22.5 × 105 – 1.68 × 105 = 20.8 × 105 = 2.08 × 106 J