50 {gm} ice at -10^{\circ} {C} is mixed with 20 {gm} steam at 100^{\circ} {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^{\circ} {C}, Take latent heat of vaporization of water =540 cal\/gm, Latent heat of fusion of water =80 callgm, Specific heat capacity of water =1 {cal} \/ {gm}-{ }^{\circ} {C}, Specific heat capacity of ice = 0.5 {cal} \/ {gm}-{ }^{\circ} {C}]

Mass of water remaining is: 67.4 {gm}

Mass of water remaining is 67.87 {gm}

Mass of steam remaining is 2.6 {gm}

Mass of steam remaining is 2.13 {gm}

JEE Advanced 2020 Concept Recapitulation Test 4 Paper 2

Question : 7 of 54

Marks: +1, -0

50gm ice at −10∘C is mixed with 20gm steam at 100∘C. When the mixture finally reaches its steady state inside a calorimeter of water equivalent 1.5gm 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 callgm, Specific heat capacity of water =1cal∕gm−‌∘C, Specific heat capacity of ice = 0.5cal∕gm−‌∘C]

Mass of water remaining is: 67.4gm
Mass of water remaining is 67.87gm
Mass of steam remaining is 2.6gm
Mass of steam remaining is 2.13gm

Validate

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