Practice Energy

Calculation Quiz

How much energy does it take to convert 722 grams of ice at 211C to steam at 675C?(Be sure to draw and label the appropriate heating or cooling curve.) Provided information: heat of fusion = 6.0 kJ/molheat of vaporization = 40.7 kJ/molspecific heat capacity of ice = 2.1 J/g∙Cspecific heat capacity of steam = 1.8 J/g∙C

Label the graphs with the correct

temperatures!

-400

-200

0

200

400

600

800

-211

00

100 100

675

Heat Added

Tem

per

atu

re

Step 1: Convert the mass in grams to moles.

mole H2O 1

g H2O 18.0

g H2O = 40.1 mol H2O

722

Step 2 Heat the ice from 211C to its melting

point of 0C.

-400

-200

0

200

400

600

800

-211

00

100 100

675

Heat Added

Tem

per

atu

re

q = mcΔTq = (722 g)(2.1 J/g∙C)(0 (211C))

q = 3.20 x 105 J

Step 3 Convert ice

to liquid water - (melt

the ice!!) -400

-200

0

200

400

600

800

-211

00

100 100

675

Heat Added

Tem

per

atu

re

q = ΔHfusion ∙molesq = (6.0 kJ/mol)(40.1 mol)

q = 241 kJ = 241,000 J

Step 4 Heat the ice from 0C to its boiling point of 100C.

-400

-200

0

200

400

600

800

-211

00

100 100

675

Heat Added

Tem

per

atu

re

q = mcΔTq = (722 g)(4.18 J/g∙C)(100 0C)

q = 302,000 J

Step 5 Convert water to

steam - (boil the water!!)

-400

-200

0

200

400

600

800

-211

00

100 100

675

Heat Added

Tem

per

atu

re

q = ΔHvaporization ∙molesq = (40.7 kJ/mol)(40.1 mol)

q = 1630 kJ = 1,630,000 J

Step 6 Heat the

steam from 100C to 675C.

-400

-200

0

200

400

600

800

-211

00

100 100

675

Heat Added

Tem

per

atu

re

q = mcΔTq = (722 g)(1.8 J/g∙C)(675 100C)

q = 747,000J

Step 7: Add the heats!

qtotal = q2 + q3 + q4 + q5 + q6

qtotal = 3,240,000 J or 3.24 x 106 J

qtotal = 3,240 kJ or 3.24 x 103 kJ

qtotal = 3.2 x 105 J + 241, 000 J + 302,000 J + 1, 630,000 J + 747,000 J