Question

50 gm ice at = 10°C is mixed with 20 gm steam at 100°C. When the mixture finally reaches its steady state inside a calorimeter of water equivalent 1.5 gm then: [Assume calorimeter was initially at 0°C, take latent heat of vaporization of water = 1 cal/gm-°C, specific heat capacity of ice = 0.5 cal/gm°C] 

(A) Mass of water remaining is: 67.4 gm 

(B) Mass of water remaining is: 67.87 gm 

(C) Mass of steam remaining is: 2.6 gm 

(D) Mass of steam remaining is: 2.13 gm

Answer 1

Correct option  (A, C)

Explanation:

∴ Heat absorbed by ice and calorimeter to reach 100° C water

= 250 + 4000 + 5000 + 150 = 9400 cal

∴ Amount of steam converted into water 

∴ Amount of water remaining = 50 + 17.40 = 67.4 gm 

∴ Amount of steam remaining = 20 – 17.4 = 2.6 gm