Question

A thermally insulated vessel contains $150\,g$ of water at $0^{\circ}C$. Then the air from the vessel is pumped out adiabatically. A fraction of water turns into ice and the rest evaporates at $0^{\circ}C$ itself. The mass of evaporated water will be closest to : (Latent heat of vaporization of water $= 2.10 \times 10^6 \; J \; kg^{-1}$ and Latent heat of Fusion of water = $3.36 \times 10^5 \; J \; kg^{-1}$)

• A. 130 g
• B. 35 g
• C. 20 g
• D. 150 g

Answer 1

Answer: (C)

20 g

Answer 2

Suppose'm' gram o f water evaporates then, heat required
$\Delta Q_{req} = mL_V$
Mass that converts into ice = (150 - m)
So, heat released in this process
$\Delta Q_{rel} = (150 - m) L_f$
Now,
$\Delta Q_{rel} = \Delta Q_{req}$
$(150 - m ) L_f =mL_V$
$m(L_f + L_v) = 150 L_f$
$m = \frac{150 L_f}{L_f + L_v}$
$m = 20 \,g$