Please help, and faster response with right work will get all the points 100 g o

Question

Please help, and faster response with right work will get all the points

100 g of an ice chunk at -5 degree C is in a thermally insulated container. We want to turn it into water vapor at its boiling point. The specific heat of ice is 2220 J/kg-K, that of water is 4190 J/kg-K. Heat of fusion for water is 333 kJ/kg and heat of vaporization is 2256 kJ/kg. Find the amount of heat required to turn the ice into water at 0 degree C. Find the amount of heat required to turn water at 0 degree C to water at 100 degree C. Find the amount of heat required to turn water at 100 degree C to steam at the same temperature. Find the entropy change of process (1). Find the entropy change of process (2).

Explanation / Answer

mass=0.1 kg

initial temperature=-5 degree celcius=268 K

1)amount of heat required to turn the ice into water at 0 degree=0.1*2220*(0-(-5))+333*10^3*0.1=34410 J

2)

amount of heat to turn water at 0 degree to water at 100 degree=0.1*4190*100=41900 J

3)to turn at 100 degree at same temeprature=0.1*2256*10^3=225600 J

4)entropy change of process 1=heat/temeprature=34410/273=126.04 J/K

5)entropy change of process 2=heat/temperature=41900/373=112.334 J/K

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