Calculate the ratio of the amount of heat energy lost during the conversion of 20 g of water at `0^(@)C` to ice at `0^(@)C` to heat energy gained during the conversion of same amount of water at `100^(@)C` to steam at `100^(@)C`.

Explanation

Latent heat of fusion or freezing is 80 cal/g, i.e., 1g of ice requires 80 cal.
20 g ice requires? = 20 + 80 = 1600 cal
Latent heat of vaporisation of water is 540 kg cal/g,
i.e., 1 g water requires 540 cal.
20 g water requires? = 20 `xx` 540 = 10,800 cal
Ratio = 1600 : 10800 = 1 : 6.75