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Question

How much heat must be removed by a refrigerator from 4 kg of water at 70C to convert it to ice cube at 10C ?
[Take specific heat of water Cw=4200 J/kg/C, Latent heat of fusion of iceLf=334000 J/kg, Specific heat of ice Ci=2100 J/kg K]
  1. 3521 kJ
  2. 4200 kJ
  3. 2596 kJ
  4. 2352 kJ


Solution

The correct option is C 2596 kJ
Energy removed when temperature of water changes from 70C to 0C
Q1=mcwΔT
=4×4200×(700)
=4×4200×70=1176000 J
Energy removed to freeze 4 kg of water.
Q2=mLw=4×334000
Q2=1336000 J
Energy to be removed to reduce the temperature of ice from 0C to 10C
Q3=mciΔ T
=4×2100×(0(10)
=4×2100×10=84000 J
Therefore, total heat removed
=Q1+Q2+Q3=1176000+1336000+84000
=2596000 J=2596 kJ
Hence, option (b) is correct. 

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