A 75-g ice cube at 0°C is heated until 68.4 g has become water at 100°C...
A 75-g ice cube at 0°C is heated until 68.4 g has become water at 100°C and 6.6 g has become steam at 100°C. How much energy was added to accomplish the transformation? kJ
Homework Answers
heat added = heat required to melt 68.4 g of ice + heat
required to convert 6.6 g to steam
heat required for 68.4 of ice to melt to water at 100 Q1 = (Mw*L)+(Mw*Sw*dT)
Lf = 334 J/g
Sw = 4.186 J/g
Lv = 2230 J/g
Q1 = (68.4*334) + (68.4*4.186*100)
Q1 = 51477.84 J
heat required to convert 6.6 g to steam Q2 = Msteam*( Lf +
(Sw*dT) + Lv )
Q2 = 6.6*( 334 + (4.186*100) + 2230 )
Q2 = 19685.16 J
Q = Q1 +Q2
Q = 51477.84 + 19685.16 = 71163 J
Q = 71.1 KJ
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