Assuming density of water = 1,
specific heat = 1 cal/g and
initial temperature of ice = 0°C
latent heat of fusion of ice
mass of ice = x
mass of water = 350
calories above 0° before mixing:
E0 = 350*24+x*0+x*(-80)
calories after mixing:
E1 = (350+x)*4
Equating calories before & after mixing:
350*24-80x=(350+x)*4
Solve for x to get x=250/3 g
How much ice (in grams) would have to melt to lower the temperature of 350 of water from 24 to 4? (Assume the density of water is 1.0 .)
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