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Work done in converting 1g of ice at -10°C into steam at 100°C is 3.04 kJ 6.05 kJ 0.721 kJ 0.616 kJ

Work done in converting 1g of ice at -10°C into steam at 100°C is
  • 3.04 kJ
  • 6.05 kJ
  • 0.721 kJ
  • 0.616 kJ

Grade:12

1 Answers

Arun
25763 Points
11 months ago
Heat Energy Required to convert 1 g ice into 1 g water = mL
[L is the Latent heat of melting of Ice.]
= 1g × 336 J/g.
= 336 J.
Heat Energy Required to raise the temperature of the Water from 0°  C to 100°C = mcΔt
[c is the specific heat capacity of water = 4200 J/kg°C.]
= 1/1000 × 4200 × 100
= 420 J.
Heat Energy Required to convert water at 100° C into steam = mL
[L is the latent heat of Vaporization of Water = 2280 J/g.]
= 1 × 2260 J/g.
= 2260 J
Total Energy Required in this Thermodynamic Process = 336 + 420 + 2260
= 3016 J.
Hope it helps.
 

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