Question

A thermally insulated vessel contains some water at $0^{\circ }\text{C}$. The vessel is connected to a vacuum pump to pump out water vapour. This results in some water getting frozen. It is given that the latent heat of vaporization of water at $0^{\circ }\text{C}$ is $21\times {10}^5\text{J/kg}$ and latent heat of freezing of water at $0^{\circ }\text{C}$ is $3.36\times {10}^5\text{J/kg}$. The maximum percentage of water vapour that will be solidified in this manner will be -

• A. $86.2\%$
• B. $33.6\%$
• C. $21\%$
• D. $24.36\%$

Answer 1

The correct option is A $86.2\%$

As the system is thermally isolated, the heat required for vaporization is extracted from the water itself. Since the water in the vessel is at $0^{\circ }\text{C}$, this results in part of the water getting frozen. Since the vapor is pumped out, the process will continue till all the remaining water is converted into ice.

Let $x$ be the maximum fraction of water that solidifies and $(1-x)$ is the fraction that gets vaporized.
By using,
$m_i{L}_f=m_s{l}_v$ at $0^{\circ }\text{C}$, we can say that,
$x\times 3.36\times {10}^5=(1-x)\times 21\times {10}^5$
$\Rightarrow 3.36x+21x=21$
or $x=0.862$ or $86.2\%$
Thus, option (a) is the correct answer.