A thermally insulated pot has 150 g ice at temperature ` 0^(@)C`. How much steam of ` 100^(@) C ` has to be mixed to it , so that water of temperature ` 50^(@)C` will be obtained ? ( Given : Latent heat of melting of ice = 80 cal/g, latent of vaporization of water = 540 cal/g, spcific heat of water = ` 1 cal// g.^(@)C)`

Explanation

Data `: m_(1) = 150 g,Delta T_(1) = 50^(@)C - 0^(@) C =50^(@)C, c_(w) = 1 cal //g .^(@)C , L_(1) = 80 cal//g,L _(2) = 540 cal //g , Delta T_(2) = 100^(@) C - 50^(@)C = 50^(@) C, m_(2) =?`
`Q_(1)` ( heat absorbed by ice ) `m_(1) L_(1) =150 gxx80 cal //g = 12000 cal`
`Q_(2) ` ( heat absorbed by water formed on melting of ice ) `= m_(1) c_(w) Delta T_(1)`
`= 150 g xx 1 cal //g .^(@) C xx 50 ^(@)C = 7500 cal `
`Q_(3) `( heat given out by steam ) `= m_(2) L_(2) = m_(2) xx 540 cal //g`
`Q_(4)` ( heat given out by water formed on condensation of steam )
`m_(2) c_(w) Delta T_(2) = m_(2) xx 1 cal //g. ^(@) C xx 50^(@) C `
According to the principle of heat exchange, `Q_(1) + Q_(2) = Q_(3) + Q_(4)`
`:. 12000 cal + 7500 cal =m_(2)xx 540 cal //g + m_(2) xx 50 cal //g `
`:. 19500 cal =m_(2) ( 540 + 50) cal //g `
`:. m_(2) = ( 19500)/(590) g =33.05 g `
`33.05 g ` of steam is to be mixed.