Heat and Thermodynamics 1 Question 2
2. A thermally insulated vessel contains $150 g$ of water at $0^{\circ} C$. Then, the air from the vessel is pumped out adiabatically. A fraction of water turns into ice and the rest evaporates at $0^{\circ} C$ itself. The mass of evaporated water will be closest to
(Latent heat of vaporisation of water $=2.10 \times 10^{6} Jkg^{-1}$ and latent heat of fusion of water $=3.36 \times 10^{5} Jkg^{-1}$ )
(2019 Main, 8 April I)
(a) $150 g$
(b) $20 g$
(c) $130 g$
(d) $35 g$
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Solution:
- Let $x$ grams of water is evaporated.
According to the principle of calorimetry,
Heat lost by freezing water (that turns into ice) $=$ Heat gained by evaporated water Given, mass of water $=150 g$
$$ \begin{aligned} & \Rightarrow(150-x) \times 10^{-3} \times 3.36 \times 10^{5} \\ & =x \times 10^{-3} \times 2.10 \times 10^{6} \\ & \Rightarrow \quad(150-x) \times 3.36=21 x \\ & \Rightarrow \quad x=\frac{150}{7.25}=20.6 \\ & \therefore \quad x \approx 20 g \end{aligned} $$
