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1 kg of ice at 0 0 C is mixed with 1 kg of steam at 100 0 C. What will be the composition of the system when thermal equilibrium is reached? Latent heat of fussion of ice=3.36*10 5 J/kg and latent heat of vaporization of water=2.26*10 6 J/kg.

1 kg of ice at 00C is mixed with 1 kg of steam at 1000C. What will be the composition of the system when thermal equilibrium is reached? Latent heat of fussion of ice=3.36*105 J/kg and latent heat of vaporization of water=2.26*106 J/kg.

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3 Answers

Babulal Prasad
14 Points
11 years ago
heat needed to melt the ice = mL = 1* 3.36*100000 =336000 J heat needed to increase the temperature of water to 100 degree centigrade=1*4200=4200 J total heat requirement= 340200 suppose m mass of steam is converted in to water then m*2260000= 340200 m= 340200/2260000=150.5 g final composition water=1150.5 g & steam=849.5 g
Vidya Iyer
15 Points
4 years ago
Energy required to melt ice completely= 1000*80 cal
Energy required to raise it's temperature to 100°= 1000*1*100
Total energy=1000(80+100) cal
Mass of steam required= m
m*540*1000 =180*1000
m= 0.33kg
Kushagra Madhukar
askIITians Faculty 628 Points
2 years ago
Dear student,
Please find the solution to your problem below.
 
Heat needed to melt the ice = mL = 1* 3.36*100000 =336000 J
Heat needed to increase the temperature of water to 100 degree centigrade=1*4200=4200 J
Total heat requirement= 340200 J
Suppose m mass of steam is converted in to water
then, m*2260000= 340200
m = 340200/2260000 = 150.5 g
Final composition water = 1150.5 g & steam=849.5 g
 
Thanks and regards,
Kushagra

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