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heat lost by the water = heat gained by ice for changing into water and rising the temp. to 20degree celsius
100*1*(30-20) = m*80 + m*1*(20-0)
solve it
specific heat capacity of water = 1 cal / g ; Tf = 20 ; Ti = 30 ;
Heat lost by water = m * s * (Tf - Ti )
= 100 * 1 cal/g * ( 20 - 30 ) = -1000 J
Heat required to melt ice = m L = m * 80
Heat required to raise temperature of ice water to 20 = m * 1 cal/g * 10
heat lost = heat gained
-1000 = -m * 80 + 10 * m
-1000 = -90 * m
mass of ice (m) = 1000/90 = 100/9 = 11.11 grams.
Therefore 1.11 grams of ice is required to change water from 30 degrees to 20 degree celcius.
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