(a) Mass of ice initially in the calorimeter = 250 g - 10 g = 240 g
(b) Rise in temperature = (45 J/K x (25 g + 15 g)) / (4200 J/kg.K x (250 g + 25 g))
= 0.068 K
Dry saturated steam at 100 degree C is passed into 250 gm of a mixture of ice and water contained in a calorimeter of thermal capacity 45J/K.When all the ice has just melted,the mass of the content has increased by 10 gm due to condensed steam.Assuming no heat exchange between the calorimeter and the surroundings,calculate
(a) the mass of ice initially in the calorimeter,and
(b) the rise in temperature when the passage of steam is continued until a further 15 gm of steam has condensed and the mixture is in thermal equilibrium.Specific latent heat of vaporization of water is 2260000 J/kg and specific heat capacity of water is 4200J/kg.K.
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