JEE PYQ: Thermodynamics Question 32
Question 32 - 2019 (08 Apr Shift 1)
A thermally insulated vessel contains 150 g 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 closed to :
(Latent heat of vaporization of water $= 2.10 \times 10^6$ J kg$^{-1}$ and Latent heat of Fusion of water $= 3.36 \times 10^5$ J kg$^{-1}$)
(1) 150 g
(2) 20 g
(3) 130 g
(4) 35 g
Show Answer
Answer: (2)
Solution
Suppose amount of water evaporated be M gram. Then $(150 - M)$ gram water converted into ice. So, heat consumed in evaporation = Heat released in fusion.
$M \times L_v = (150 - M) \times L_f$
$M \times 2.1 \times 10^6 = (150 - M) \times 3.36 \times 10^5$
$\Rightarrow M \approx 20$ g
BETA
