JEE Mains · Physics · STD 11 - 10.1, thermonetry,thermal expansion and calorimetry
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 \( = 3.36 \times10^5\,Jkg^{-1}\) )
- A \(35\)
- B \(150\)
- C \(130\)
- D \(20\)
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, \(\Delta {Q_{rel}} = \Delta {Q_{req}}\)…
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