Suppose m grams of water evaporate. Then, heat required
ΔQreq=mLv
Mass that converts into ice =(150−m)
Heat released in the process,
ΔQrel=(150−m)Lf
Now, ΔQrel=ΔQreq
(150−m)Lf=mLv
m(Lf+Lv)=150Lf
m=Lf+Lv150Lf
m=20g
JEE Main 2019 — Physics Thermodynamics
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 =2.10×106Jkg−1 and Latent heat of Fusion of water =3.36×105Jkg−1 )
Held on 8 Apr 2019 · Verified 6 Jul 2026.
35g
20g
130g
150g
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