50 gm ice at -10°C is mixed with 10gm 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,

Question

50 gm ice at -10°C is mixed with 10gm 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 = 540 cal / gm , Latent heat of fusion of water = 80 cal / gm and specific heat capacity of water = 1 cal / gm − ∘ C , specific heat of ice = 0.5 cal / gm ∘ C ]

TestHub · Accepted Answer

Heating 50 g ice from -10°C then melting it requires 250+4000=4250 cal.

Condensing 10 g steam to water at 0°C releases 6400 cal, leaving all ice melted and final temperature between 0 and 100°C; solving gives ice remainder about 3 g under the paper setup.

Thermal Properties of Matter - Calorimetry Question with Solution | TestHub

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