practice energy calculation quiz
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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
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