How much heat (in kJ) is required to warm 13.0 g of ice, initially at -13.0 ∘C, to steam at 112.0 ∘C? The heat capacity of ice is 2.09 J/g⋅∘C and that of steam is 2.01 J/g⋅∘C, the heat of fusion for water is 6.02 kJ/mol, and the heat of vaporization for water is 40.7 kJ/mol.
+33653 "How much heat (in kJ) is required to warm 13.0 g of ice, initially at -13.0 ∘C, to steam at 112.0 ∘C? The heat capacity of ice is 2.09 J/g⋅∘C and that of steam is 2.01 J/g⋅∘C, the heat of fusion for water is 6.02 kJ/mol, and the heat of vaporization for water is 40.7 kJ/mol."
Latent heat of fusion = 334 kJ/kg Latent heat of vaporisation = 2264.76 kJ/kg
Specific heat capacity of water = 4.18 J/(gK)
Ice from -13 to 0°C E1 = 13*2.09*13 J → 353.21 J → 0.353 kJ
Ice to water at 0°C E2 = 0.013*334 kJ → 4.342 kJ (Note: 13g = 0.013 kg)
Water at 0 to 100°C E3 = 13*4.18*100 J → 5434 J → 5.434 kJ
Water to steam at 100°C E4 = 0.013*2264.76 kJ → 29.442 kJ
Steam from 100 to 112°C E5 = 13*2.01*12 J → 313.56 J → 0.314 kJ
I'll leave you to add up the five energies!
.
