If 80 g steam of temperature `97^@C` is released on an ice slab of temperature `0^@C` ,how much ice will melt?How much energy will be transferred to water? Given : specific heat capacity of water, c = 1 cal/g °C Latent heat of melting of ice = 80 cal/g Latent heat of vaporization of water = 540 cal/g

Explanation

Latent heat of melting the `ice = L_("melt") = 80 cal//g`
Latent heat of vaporization of water `= L_(vap), = 540 cal//g`
Solution: mass of steam = `m_("steam") = 80 g`
Temperature of steam = `97" "^(@)C`
Temperature of ice = `T_(ice) = 0" "^(@)C`
Heat released during conversion of steam of temperature `97" "^(@)C` into water of
temperature `97" "^(@)C = m_("steam") xx L_(vap)`
= `80 xx (97 - 0) xx 1 = 80 xx 97` ------------- (2)
Total heat gained by the ice `80 xx 540 + 97` from equations (1) and (2)
Some mass, of the ice, `m_(ice)` will melt due to this heat gained by the ice, then,
`m_(ice) XL_("melt") = 50960` cal
`m_(ice) = 637 g`
Thus, 637 g ice will melt and 50960 cal kcal will be given to the ice.