How many grams of ice at -17°C must be added to 649 grams of water that is initi

Question

How many grams of ice at -17°C must be added to 649 grams of water that is initially at a temperature of 64°C to produce water at a final temperature of 11°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 ? °C and of ice is 2000 J/kg ? °C. For water the normal melting point is 0°C and the heat of fusion is 334 it × 103 J/kg. The normal boiling point is 100° C and the heat of vaporization is 2.256 × 106 J/kg.

Explanation / Answer

Q out of water = Q into ice So m*c*?T)water = m*c*?T)ice + m*Lf + m*c*?T Now 170g*4.19J/g-oC*(92 - 6) = m*(2.00*21 + 334 + 4.19*(6)) So m = 170*4.19*86 = m*(401.14) So m = 170*4.19*86/401.14 = 153g

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