Question

Calculate the amount of energy in kilojoules needed to change 279g of water ice at ?10 ?C to steam at 125 ?C. The following constants may be useful:

Cm (ice)=36.57 J/(mol??C)

Cm (water)=75.40 J/(mol??C)

Cm (steam)=36.04 J/(mol??C)

?Hfus=+6.01 kJ/mol

?Hvap=+40.67 kJ/mol

Answer 1

Moles of water = 279 g / 18 g/mol = 15.5 mol

Step 1:

Heat required to convert 279 g of ice at -10 oC to 0 oC ice

Q1 = mcdT

Q1 = 15.5 x 36.57 x 10

Q1 =  5668.35 J = 5.668 kJ

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Step 2:

Heat required to convert 279 g of ice at 0 oC ice to 0 oC water

Q2 = m x delta Hfus

Q2 = 15.5 x 6.01 = 93.15 kJ

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Step 3:

Heat required to convert 279 g of water at 0 oC to water at 100 oC

Q3 = mcdT

Q3 = 15.5 x 75.40 x 100

Q3 = 116870 J = 116.87 kJ

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Step 4:

Heat required to convert 279 g of water at 100 oC to steam at 100 oC

Q4 = m x delta Hvap

Q4 = 15.5 x 40.67

Q4 =  630.38 kJ

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Step 5:

Heat required to convert 279 g of steam at 100 oC to steam at 125 oC

Q5 = mcdT

Q5 = 15.5 x 36.04 x 25

Q5 =  13965.5 J = 13.96 kJ

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Total energy = Q1 + Q2 + Q3 + Q4 + Q5

= 5.668 kJ + 93.15 kJ + 116.87 kJ + 630.38 kJ + 13.96 kJ

= 860.03 kJ