Physical idea

The vessel is thermally insulated, so no heat is exchanged with surroundings.

When air is pumped out adiabatically, some water evaporates at $0^\circ\text{C}$, requiring latent heat of vaporization. Since no external heat is available, this heat must come from the water itself.

Therefore, another part of the water freezes at $0^\circ\text{C}$ and releases latent heat of fusion.

So, in equilibrium: $\text{heat released by freezing} = \text{heat absorbed in evaporation}$

Let masses be

Let

• mass of water frozen $= m_f$
• mass of water evaporated $= m_e$

Initial mass of water $=150\,\text{g}=0.150\,\text{kg}$.

Apply energy balance

Heat released by freezing: $Q_{\text{freeze}} = m_f L_f$

Heat absorbed by evaporation: $Q_{\text{evap}} = m_e L_v$

Hence, $m_f L_f = m_e L_v$

Using $L_f = 3.36\times10^5\,\text{J kg}^{-1}, \qquad L_v = 2.10\times10^6\,\text{J kg}^{-1}$

$m_f(3.36\times10^5)=m_e(2.10\times10^6)$

$m_f = m_e\left(\frac{2.10\times10^6}{3.36\times10^5}\right)=m_e\left(\frac{210}{33.6}\right)=6.25m_e$

Use mass conservation

Since finally some becomes ice and the rest evaporates, all the initial water is accounted for by these two parts: $m_f + m_e = 0.150$

Substitute $m_f=6.25m_e$: $6.25m_e + m_e = 0.150$ $7.25m_e = 0.150$ $m_e = \frac{0.150}{7.25} \approx 0.0207\,\text{kg}$

$m_e \approx 20.7\,\text{g}$

Closest option

$m_e \approx 21\,\text{g}$

Closest option is 20 g.

Answer check with stored answer

Derived answer: Option C (20 g).

Stored correct answer: C.

They match.