Using the heat of fusion for water, 334 J/g, the heat of vaporization for water,

Question

Using the heat of fusion for water, 334 J/g, the heat of vaporization for water, 2260 J/g, and the specific heat of water, 4.184 J/g?C, calculate the total amount of heat for each of the following:

joules needed to warm 14.5 g of water at 19 ?C to 92 ?C

joules needed to melt 59.0 g of ice at 0.0 ?C and to warm the liquid to 53.5 ?C

kilojoules released when 20.0 g of steam condenses at 100 ?C and the liquid cools to 0 ?C

kilojoules needed to melt 26.0 g of ice at 0 ?C, warm the liquid to 100 ?C, and change it to steam at 100 ?C

Explanation / Answer

14.5 g of water at 19 C to 92 C

m = 14.5 g Cp = 4.184 J/g.C T2 = 92 C T1 = 19 C

energy = m * Cp * (T2 - T1) = 14.5 g * 4.184 J/g.C * ( 92 - 19) C = 4428.764 J = 4.43 kJ

59 g of ice at 0 C to liquid at 53.5 C

m = 59 g

T1 = 0 C

T2 = 53.5 C

Cp = 4.184

heat of fusion = 334 J/g

Energy = 334 J/g * 59 g + 59 g * 4.184 J/g.C *(53.5 - 0 ) C = 19706 J + 13207.8 J = 32912.8 J

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