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 = 540 cal g⁻¹, Latent heat of fusion of water = 80 cal g⁻¹, specific heat capacity of water = 1 cal g⁻¹ °C⁻¹, specific heat capacity of ice = 0.5 cal g⁻¹ °C⁻¹]
Physics · Thermodynamics
JEE Advanced

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 = 540 cal g⁻¹, Latent heat of fusion of water = 80 cal g⁻¹, specific heat capacity of water = 1 cal g⁻¹ °C⁻¹, specific heat capacity of ice = 0.5 cal g⁻¹ °C⁻¹]

Physics · Thermodynamics

G1 Physics · Accepted Answer

Apply energy conservation. Sensible heat is Q=mcΔT and phase-change heat is Q=mL. Heat lost by hotter parts equals heat gained by colder parts. Evaluating the given data/statements gives A: Mass of water remaining is : 67.4 gm; B: Mass of steam remaining is : 2.6 gm. Therefore the correct answer is A, B.