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JEE Main 20198 AprilMorning Shift
Exam
JEE Main
Year
2019
Date
8 April
Shift
Morning Shift
Subject
Physics
Chapter
Thermal Properties of Matter

JEE Main 2019 · 8 April · Morning Shift · Physics · Thermal Properties of Matter

A thermally insulated vessel contains 150g of water at 0 ° C . Then...

A thermally insulated vessel contains 150g150g of water at 0°C0 ° C . Then the air from the vessel is pumped out adiabatically. A fraction of water turns into ice and the rest evaporates at 0°C0 ° C itself. The mass of evaporated water will be closest to: (Latent heat of vaporization of water =2.10×106Jkg−1= 2.10 \times {10}^{6}Jk{g}^{- 1} and Latent heat of Fusion of water =3.36×105Jkg−1= 3.36 \times {10}^{5}Jk{g}^{- 1} )

Options

  1. A35g35g
  2. B20g20g
  3. C130g130g
  4. D150g150g

Correct answer

B. 20g20g

Step-by-step solution

Suppose mm grams of water evaporate. Then, heat required
ΔQreq=mLv\Delta{Q}_{req} = m{L}_{v}
Mass that converts into ice =(150−m)<br/>Heatreleased= \left(150 - m\right) <br/>Heat released in the process,
\Delta{Q}_{rel} = \left(150 - m\right)Lf{L}_{f}
Now, ΔQrel=ΔQreq\Delta{Q}_{rel} = \Delta{Q}_{req}
\left(150 - m\right)Lf=mLv{L}_{f} = m{L}_{v}
m(Lf+Lv)=150Lfm\left({L}_{f} + {L}_{v}\right) = 150{L}_{f}
m=150LfLf+Lvm = \dfrac{150{L}_{f}}{{L}_{f} + {L}_{v}}
m=20gm = 20g
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