What mass of superheated steam at 120 C must be added to 1 kg of ice at 0 degree

Question

What mass of superheated steam at 120 C must be added to 1 kg of ice at 0 degrees C to obtain liquid water at 20 degrees C? Sp. heat of steam= 0.48 cal/gm C, sp. heat of water = 1.0 cal/gm C, heat of fusion of ice = 80 cal/gm, heat of condensation of steam = 540 cal/gm C What mass of superheated steam at 120 C must be added to 1 kg of ice at 0 degrees C to obtain liquid water at 20 degrees C? Sp. heat of steam= 0.48 cal/gm C, sp. heat of water = 1.0 cal/gm C, heat of fusion of ice = 80 cal/gm, heat of condensation of steam = 540 cal/gm C

Explanation / Answer

heat lose = heat gain

m*0.48*(120-100) + m*540 + m*1*(100 - 20) = 1000*80 + 1000*1*20

629.6*m = 100000

m = 100000/629.6 = 158.83 gm

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