Question

A thermally insulated vessel contains some water at 0°C. The vessel is connected to a vacuum pump to pump out water vapour. This results in some water getting frozen. It is given Latent heat of vaporization of water at 0°C = 21 × 10⁵ J/kg and latent heat of freezing of water = 3.36 × 10⁵ J/kg. The maximum percentage amount of water that will be solidified in this manner will be

• A. 86.2%
• B. 33.6%
• C. 21%
• D. 24.26%

Answer 1

Answer: (A)

86.2%

Answer 2

In a thermally insulated vessel, the heat released by the freezing of water must be equal to the heat absorbed by the evaporation of water.

Let $m$ be the initial mass of water. Let $m_f$ be the mass of water that freezes. Let $m_e$ be the mass of water that evaporates.

Heat released by freezing: $Q_{freeze} = m_f L_f$ Heat absorbed by evaporation: $Q_{evap} = m_e L_v$

Where $L_f$ is the latent heat of freezing and $L_v$ is the latent heat of vaporization.

For the process to continue until the maximum amount of water is solidified, the heat released by freezing must equal the heat absorbed by evaporation: $m_f L_f = m_e L_v$

The total mass of water is conserved, so $m = m_f + m_e$. Therefore, $m_e = m - m_f$.

Substituting this into the energy balance equation: $m_f L_f = (m - m_f) L_v$

We want to find the maximum percentage of water solidified, which is $\frac{m_f}{m} \times 100\%$. Let $f = \frac{m_f}{m}$. $f m L_f = (m - f m) L_v$ Divide both sides by $m$: $f L_f = (1 - f) L_v$ $f L_f = L_v - f L_v$ $f L_f + f L_v = L_v$ $f (L_f + L_v) = L_v$ $f = \frac{L_v}{L_f + L_v}$

Given values: $L_v = 21 \times 10^5$ J/kg $L_f = 3.36 \times 10^5$ J/kg

$f = \frac{21 \times 10^5}{3.36 \times 10^5 + 21 \times 10^5} = \frac{21}{3.36 + 21} = \frac{21}{24.36}$

Percentage solidified $= f \times 100\% = \frac{21}{24.36} \times 100\% \approx 86.20689655\%$

Rounding to one decimal place, the maximum percentage of water that will be solidified is 86.2%.