Question

Use these relationships to determine the number of calories to change 1.5 kg of 0∘c ice water to 1.5 kg of 100∘c boiling water.

Answer 1

The latent heat of fusion of water is 80 cal/g.
The specific heat of water is 1 cal/g-C.
The latent heat of vaporization of water is 540 cal/g.
Therefore, if we have 1.5 kg = 1500 g, the total heat requirement is:
1500 g[(80 cal/g) + (1 cal/g-C)(100 - 0)C + (540 cal/g)] = 1500 g(720 cal/g) = 1,080,000 cal.