Question

If water at $0^{\circ }C$, kept in a container with an open top, is placed in a large evacuated chamber, if the specific latent heat of vaporization of water at $0^{\circ }C$ is $\eta$ times the specific latent heat of freezing of water at $0^{\circ }C$, the fraction of water that will ultimately freeze is

• A. $\frac{\eta }{\eta +1}$
• B. $\frac{\eta -1}{\eta +1}$
• C. $\frac{\eta -1}{\eta }$
• D. $\frac{1}{\eta }$

Answer 1

The correct option is A $\frac{\eta }{\eta +1}$

We know that: 𝑄=𝑚𝐿
Given: $L_{steam}=\eta L_{ice}$

Let $m$ be mass of water that converted into ice and $m^{\prime }$ be mass that converted into steam.

If water at $0^{\circ }C$, kept in a container with an open top, is placed in a large evacuated chamber, part of the water will vaporize and the rest will freeze.

This is because portion of water losing heat to freeze will be used by the other portion of water to vaporise.

$m^{\prime }{L}_{steam}=m{L}_{ice}$

$m^{\prime }\eta L_{ice}=m{L}_{ice}$

$m^{\prime }=\frac{m}{\eta }\dots \left(i\right)$

Then fraction of mass that converted into ice is given by

From equation $\left(i\right)$

$\frac{m}{m+m^{\prime }}=\frac{m}{m+\frac{m}{\eta }}=\frac{\eta }{\eta +1}$

FINAL ANSWER: (b).