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If steam of mass 100g and temperature 10...

If steam of mass 100g and temperature `100^(@)C ` is released on an ice slab of temperature `0^(@)C ` , how much ice will melt ?

Text Solution

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mass of ice `m = ? , L_(2) = ` 80 cal/g ,c (water ) = ` 1 "cal/g"^(@) C `
According to the principle of heat lost by hot body = heat gained by cold body .
Conversion of steam into water :
` Q_(1) = m_(1)L_91) = 100g xx 540 "cal/g" = 54000 cal `
Decrease in the temperature of this water to ` 0^(@)C`
` Q_(2) = m_(1)cxx (T_(1) - 0^(@)C) = 100 g xx 1 "cal/g"^(@)C xx (100^(@)C - 0^(@)C) = 1000 cal`
Melting of ice = ` Q_(3) = mL_(2) = mxx 80 "cal/g"`
Now , ` Q_(1) + Q_(2) = Q_(3)`
( 54000 + 10000) cal. = ` m xx 80` cal/g
` m= ( 64000)/80 g = 800 g`
800 g of ice will melt.
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