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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 ?

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Data `:` `m_(1) = 100g, L_(1) =540 cal //g, T_(1)= 100^(@)C, ` mass of ice, `m=? , L_(2) =80 cal //g, c `( water ) `= 1 cal //g.^(@) C `
According to the principle of heat exchange , heat lost by hot body = heat gained by cold body.
Conversion of steam into water `:`
`Q_(1) = m_(1) L_(1) =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 ) =1000cal`
Melting of ice `: Q _(3) = mL_(2) `
`= m xx 80 cal//g `
Now, `Q_(1) + Q_(2) = Q_(3)`
`:. ( 5400 + 10000) cal = m xx 80 cal //g `
` :. m = ( 64000)/( 80) g = 800 g `
800 g of ice will melt.
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