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)
1. 150 g
2. 20 g
3. 130 g
4. 35 g

Answer 1

Correct Answer - Option 2 : 20 g

Concept:

Vaporisation is the phase change from liquid to vapour. Evaporation is the process of vaporisation occurring from the free surface of a liquid.

The latent heat of vapourization of a substance is defined as the heat required to change one mole of liquid at its boiling point under standard atmospheric pressure. It increases as the temperature is decreased.

The latent heat of fusion of a substance is associated with melting of a solid or freezing a liquid.

Calculation:

Let 'm' gram of water evaporates.

The heat required to convert water into water vapour is:

ΔQ_req = mL_v

Where L_v is Latent heat of vaporization of water

Mass that converts into ice = (150 – m)

The heat released during water to ice process is:

ΔQ_rel = (150 – m)L_f

Where L_f is Latent heat of Fusion of water

The energy required is equal to energy released.

ΔQ_rel = ΔQ_req

⇒ mL_v = (150 – m)L_f

⇒ mL_v = 150L_f – mL_f

⇒ mL_v + mL_f = 150L_f

⇒ m(L_v + L_f) = 150L_f

\(\Rightarrow \text{m}=\frac{150{{\text{L}}_{\text{f}}}}{\left( {{\text{L}}_{\text{v}}}+{{\text{L}}_{\text{f}}} \right)}\) 

From question, we know that,

L_f = 3.36 × 10⁵ J kg^-1

L_v = 2.10 × 10⁶ J kg^-1

On substituting the values, in the equation above,

\(\Rightarrow \text{m}=\frac{150\times \left( 3.36\times {{10}^{5}} \right)}{\left( 2.10\times {{10}^{6}} \right)+\left( 3.36\times {{10}^{5}} \right)}\) 

\(\Rightarrow \text{m}=\frac{504\times {{10}^{5}}}{\left( 21\times {{10}^{5}} \right)+\left( 3.36\times {{10}^{5}} \right)}\) 

\(\Rightarrow \text{m}=\frac{504\times {{10}^{5}}}{24.36\times {{10}^{5}}}\) 

\(\Rightarrow \text{m}=\frac{504}{24.36}=20.69\) 

Thus, the mass of evaporated water will be closest to 20 g.