Question

A thermally insulated vessel contains 150 g of water at 0°C. Then the air from the vessel is pumped out adiabatically. A fraction of water turns into ice and the rest evaporates at 0°C itself. The mass of evaporated water will be closest to ______.

(Latent heat of vaporization of water = 2.10 × 10⁶ J kg^-1 and Latent heat of Fusion of water = 3.36 × 10⁵ J kg^-1)

पर्याय

• 150 g

• 20 g

• 130 g

• 35 g

Answer 1

A thermally insulated vessel contains 150 g of water at 0°C. Then the air from the vessel is pumped out adiabatically. A fraction of water turns into ice and the rest evaporates at 0°C itself. The mass of evaporated water will be closest to 20 g.

Explanation:

Given:

A thermally insulated vessel has a total mass of water M = 150 g, if the water inside a container is T = 0°C, Water has a latent heat of vaporization of L_v = 2.10 × 10⁶ J kg^-1, Water has a latent heat of fusion with L_F = 3.36 × 10⁵ J kg^-1. 

To find: The volume of water that has evaporated after the vessel has been pushed out adiabatically. Let x g = x × 10^-3 kg be the mass of water that has evaporated.

In other words, the heat lost by freezing water (the volume of water that freezes into ice) equals the heat gained by the water that evaporates (the amount of water that turns into steam)

(150 - x) × 10^-3 × 3.36 × 10⁵ = x × 10^-3 × 2.10 × 10⁶

`150 xx 3.36 xx 10^5 = (2.10 xx 10^-6 + 3.36 xx 10^-5)x`

x = 20 g