How many grams of ice at -14 degree C must be added to 982 grams of water that i

Question

How many grams of ice at -14 degree C must be added to 982 grams of water that is initially at a temperature of 62 degree C to produce water at a final temperature of 12 degree C? Assume that no heat is lost to the surroundings and that the container has negligible mass. The specific heat of liquid water is 4190 J/kg middot degree C and of ice is 2000 J/kg middot degree C. For water the normal melting point is 0 degree C and the heat of fusion is 334 it times 10^3 J/kg. The normal boiling point is 100 degree C and the heat of vaporization is 2.256 times 10^6 J/kg.

Explanation / Answer

Apply conservation of energy

Q lost by water = Q gained ice

mc del T =  m*c*delTice + m*Lf + m*c*delT

0.982 kg * 4190J/kg-oC*(62 - 12) = m*((334000+2000*14+4190*12))

m = 0.499 kg

=499 gram

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