JEE MAIN - Physics (2019 - 8th April Morning Slot - No. 5)
A thermally insulated vessel contains 150g of
water at 0°C. Then the air from the vessel is
pumped out adiabatically. A fraction of water
turns into ice and the rest evaporates at 0°C
itself. The mass of evaporated water will be
closest to :
(Latent heat of vaporization of water
= 2.10 × 106 J kg–1 and Latent heat of Fusion
of water = 3.36 × 105 J kg–1)
35 g
130 g
20 g
150 g
Explanation
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
$$ \Rightarrow (150 - x) \times {10^{ - 3}} \times 3.36 \times {10^5} = x \times {10^{ - 3}} \times 2.10 \times {10^6}$$
$$ \Rightarrow (150 - x) \times 3.36 = 21x$$
$$ \Rightarrow x = {{150} \over {7.25}} = 20.6$$
$$\therefore$$ $$x \approx 20g$$
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