50 gm ice at − 1 0 ∘ C is mixed with 10 gm steam at 10 0 ∘ C . When the mixture finally reaches its steady state inside a calorimeter of water equivalent 1.5 gm then: [Assume calorimeter was initially

Question

50 gm ice at − 1 0 ∘ C is mixed with 10 gm steam at 10 0 ∘ 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

Warm and melt ice: Q 1 ​ = 50 × 2 1 ​ × 10 + 50 × 80 = 4250 cal Condensing steam supplies: Q 2 ​ = 10 × 540 = 5400 cal > Q 1 ​ Heat needed to reach boiling: 50 × 100 + 1.5 × 100 = 5150 cal Surplus insufficient; steam condenses. 5400 + 10 ( 100 − T ) = 4250 + 50 T + 1.5 T 2150 = 61.5 T T ≈ 3 5 ∘ C m w a t er ​ = 50 + 10 = 60 g

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