Question

The increase in internal energy of 1 kg of water at 100°c when it is converted into steam at the same temperature and at 1 atmospheric pressure will be ______.

(density of steam is 0.6 kg m^-3, latent heat of vaporisation of water = 2.25 × 10⁶ J kg^-1).

पर्याय

• 1.04 × 10⁶ J 

• 2.08 × 10⁶ J 

• 4.16 × 10⁶ J 

• 8.32 × 10⁶ J 

Answer 1

The increase in internal energy of 1 kg of water at 100°c when it is converted into steam at the same temperature and at 1 atmospheric pressure will be 2.08 × 10⁶ J.

Explanation:

Lv = 2.25 × 10⁶ J/kg

m = 1 kg

p = 1 × 10² pa

volume of water (V) = 10^-3m³

volume of steam (V) = `1/0.6`m³

Increase in volume = (`1/0.6-1/1000`) = 1.7 m³

Work done by system = P(AV) = (1 × 10⁵)(1.7)

= 1.7 × 10⁵J

ΔQ = ml_v = 2.25 × 10⁶J

change in internal energy ΔU = ΔQ - ΔW

ΔU = 2.25 × 10⁶ - 1.7 × 10⁵

= 2.25 × 10⁶ - 0.17 × 10⁶

= 2.08 × 10⁶J