Question

A vessel contains a small amount of water at $0^{\circ }\text{C}$. If the air in the vessel is rapidly pumped out, it causes freezing of the water. What percentage of the water in the container can be frozen by this method? Latent heat of vaporization and fusion are $L_v=540{\text{cal g}}^{-1}$and $L_f=80{\text{cal g}}^{-1}$ respectively.

• A. $36\%$
• B. $54\%$
• C. $87\%$
• D. $97.3\%$

Answer 1

The correct option is C $87\%$

At any temperature, there is some vapour above the surface of water. The vapour pressure depends on the amount of vapour. When the air is pumped out, the vapour gets removed and the vapour pressure falls. As a result, there is further evaporation of the liquid to saturate the air. The latent heat for evaporation is supplied by the water at $0{}^{\circ }\text{C}$ and therefore a part of it freezes.
Let $m=$ initial mass of water
$m_1$ = mass that evaporates
$m_2$ = mass that freezes
From principle of calorimetry,
Heat lost by water freezing = Heat gained by water evaporating
$m_1{L}_v=m_2{L}_f$
$\Rightarrow m_2=m_1\frac{L_v}{L_f}$
$\Rightarrow \frac{m_2}{m}=\frac{m_2}{m_1+m_2}=\frac{m_1\frac{L_v}{L_f}}{m_1+m_1\frac{L_v}{L_f}}=\frac{L_v}{L_f+L_v}$
$\therefore$ % of water that freezes by this process:
$=\frac{m_2}{m}\times 100=\frac{L_v}{L_f+L_v}\times 100$
$=\frac{540}{80+540}\times 100=87\%$
Thus, option (c) is the correct answer.