JEE Mains · Physics · STD 11 - 10.1, thermonetry,thermal expansion and calorimetry
- JEE Mains 2019
- Hard
- Numerical
A thermally insulted vessel contains \(150\, g\) of water at \(0\,^oC\). Then the air from the vessel is pumped out a adiabatically. A fraction of water turns into ice and the rest evaporates at \(0\,^oC\) itself. The mass of evaporated water will be closes to ....... \(g\) (Latent heat of vaporization of water \(= 2.10 \times10^6\, Jkg^{-1}\) and Laten heat of Fusion of water \( = 3.36 \times10^5\,Jkg^{-1}\) )
Pick an option to check your answer.
Answer & Solution
Correct Answer
(D) \(20\)
Step-by-step Solution
Detailed explanation
Suppose \('m'\) gram of water evaporates then, heat required \(\Delta {Q_{req}} = m{L_v}\)
Mass that converts into ice \(=(150-m)\)
So, heat released in this process
\(\Delta {Q_{rel}} = \left( {150 - m} \right){L_f}\)
Now,
\(\Delta {Q_{rel}} = \Delta {Q_{req}}\)
\(\left( {150 - m} \right){L_f} = m{L_v}.\)
\(M\left( {{L_f} - {L_v}} \right) = 150\,{L_f}\)
\(m = \frac{{150{L_f}}}{{{L_f} + {L_v}}}\,\,;\,\,m = 20\,g\)
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