A thermally insulated vessel contains 150, g | vedclass

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(D) Let $m$ be the mass of water that evaporates in grams.Since the process is adiabatic and no external heat is supplied,the heat released by the fre

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$20$

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(D) Let $m$ be the mass of water that evaporates in grams.Since the process is adiabatic and no external heat is supplied,the heat released by the freezing of a portion of the water must be equal to the heat absorbed by the portion that evaporates.Mass of water that turns into ice $= (150 - m)\, g$.Heat released during freezing $\Delta Q_{rel} = (150 - m) 	imes L_f$.Heat absorbed during evaporation $\Delta Q_{req} = m 	imes L_v$.Equating the two: $(150 - m) 	imes L_f = m 	imes L_v$.Substituting the values $L_f = 3.36 	imes 10^5\, J/kg$ and $L_v = 2.10 	imes 10^6\, J/kg$:$(150 - m) 	imes 3.36 	imes 10^5 = m 	imes 21.0 	imes 10^5$.$150 	imes 3.36 = m 	imes (21.0 + 3.36)$.$504 = m 	imes 24.36$.$m = 504 / 24.36 \approx 20.69\, g$.The mass of evaporated water is closest to $20\, g$.

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