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 the ice when the steam will be transformed to water?

Explanation

To find : `Q_(1) = ? , m_(1)?`
Formulae : `Q_(A) = Q_(B)`
`Q= mcDelta T, Q = mL`
Soltion :
(1) Steam
`"Steam"(97^(@)C) overset(mL)to "Water"(97^(@)C)overset(mcDeltaL)to"Water"(0^(@)C)`
`therefore Q_(A) = mL + mcDeltaT`
` = (80 xx 540) + {80 xx 1xx (97 -0)}`
` = 43200 + 7760`
`Q_(A) = 50960 cal............(i)`
(2) Ice
`Ice(0^(@)C)overset(mL)to "Water"(0^(@)C)`
`Q_(B) = m_(1),L_(1)..........(ii)`
from (i) and (ii)
`Q_(A) = Q_(B)`
`50960 = m_(1)L_(1)`
`50960 = m_(1)xx 80`
`m_(1) = (5096cancel0)/(8cancel0)`
`m_(1) = 637 g`
Hence, 637g of ice will melt and 50960 cal of heat energy is transfered to ice.