(b)
8.014
×
10
10
kJ
×
kJ
10
4.2
dynamite
ton
1
6
×
= 1.908
×
10
4
= 1.9
×
10
4
ton dynamite
The energy released by this rainstorm is equivalent to explosion of 19,000 tons of
dynamite.
5.97
Find the heat capacity of 1.7
×
10
3
gal H
2
O.
C

g
1
J
184
.
4
cm
1
g
1
L
1
cm
10
1
qt
1.057
L
1
gal
1
qt
4
O
H
gal
10
7
.
1
C
o
3
3
3
2
3
O
H
2
×
×
×
×
×
×
×
=
= 2.692
×
10
7
J/
°
C = 2.7
×
10
4
kJ/
°
C; then,
bricks
18,000
or
10
8
.
1
kg
1.8
brick
1
g
10
1
kg
1
J
0.85
C

g
1
C
1
J
10
692
.
2
4
3
o
o
7
×
=
×
×
×
×
×
Check
.
(1.7
×
~16
×
10
6
)/(~1.6
×
10
3
)
≈
17
×
10
3
bricks; the units are correct.
5.98
(a)
J
10
27
.
3
9
.
3274
C)
4
.
100
C
1
.
30
(
Cu
g
0
.
121
K
g
J
385
.
0
q
3
o
o
Cu

×
−
=
−
=
−
×
×
=
The negative sign indicates the 3.27
×
10
3
J are lost by the Cu block.
(b)
J
10
1
.
3
3138
C)
1
.
25
C
1
.
30
(
O
H
g
0
.
150
K
g
J
184
.
4
q
3
o
o
2
O
H

2
×
=
=
−
×
×
=
The positive sign indicates that 3.14
×
10
3
J are gained by the H
2
O.
5
Thermochemistry
Solutions to Exercises
127
(c)
The difference in the heat lost by the Cu and the heat gained by the water is 3.275
×
10
3
J – 3.138
×
10
3
J = 0.137
×
10
3
J = 1
×
10
2
J. The temperature change of the
calorimeter is 5.0
°
C. The heat capacity of the calorimeter in J/K is
0.137
×
10
3
J/K.
10
3
4
.
27
C
0
.
5
1
J
o
×
=
=
×
Since
O
H
2
q
is known to one decimal place, the difference has one decimal place
and the result has 1 sig fig.
If the rounded results from (a) and (b) are used,
J/K.
10
4
C
0
.
5
J
10
2
.
0
C
o
3
r
calorimete
×
=
×
=
(d)
T)
(
g
0
.
150
K
g
J
184
.
4
J
10
275
.
3
q

3
O
H
2
Δ
×
×
=
×
=
Δ
T = 5.22
°
C; T
f
= 25.1
°
C + 5.22
°
C = 30.3
°
C
5.99
(a)
From the mass of benzoic acid that produces a certain temperature change, we
can calculate the heat capacity of the calorimeter.
C
kJ/
78
.
3
7755
.
3
acid
benzoic
g
1
kJ
38
.
26
observed
change
C
642
.
1
acid
benzoic
g
235
.
0
o
o
=
=
×
Now we can use this experimentally determined heat capacity with the data for
caffeine.
caffeine
kJ/mol
10
22
.
4
caffeine
mol
1
caffeine
g
2
.
194
C
1
kJ
7755
.
3
caffeine
g
265
.
0
rise
C
525
.
1
3
o
o
×
=
×
×
(b)
The overall uncertainty is approximately equal to the sum of the uncertainties
due to each effect. The uncertainty in the mass measurement is 0.001/0.235 or
0.001/0.265, about 1 part in 235 or 1 part in 265. The uncertainty in the
temperature measurements is 0.002/1.642 or 0.002/1.525, about 1 part in 820 or 1
part in 760. Thus the uncertainty in heat of combustion from each measurement is
kJ
6
760
4220
kJ;
5
820
4220
kJ;
16
265
4220
kJ;
18
235
4220
=
=
=
=
The sum of these uncertainties is 45 kJ. In fact, the overall uncertainty is less than
this because independent errors in measurement do tend to partially cancel.
5.100
Plan
.
Use the heat capacity of H
2
O to calculate the energy required to heat the water.
Use the enthalpy of combustion of CH
4
to calculate the amount of CH
4
needed to
provide this amount of energy, assuming 100% transfer. Assume H
2
O(l) is the product
of combustion at standard conditions.
Solve.
required
kJ
272
J
10
720
.
2
C)
0
.
25
C
0
.
90
(
C

g
J
184
.
4
kg
1
g
1000
O
H
kg
00
.
1
5
o
o
o
2
=
×
=
−
×
×
×
CH
4
(g) + 2O
2
(g)
→
CO
2
(g) + 2H
2
O(l)
)
g
(
O
H
)
g
(
CH
H
O(l)
H
H
2
)
g
(
CO
H
H
2
f
4
f
2
f
2
f
rxn
o
o
o
o
o
Δ
−
Δ
−
Δ
+
Δ
=
Δ
5
Thermochemistry
Solutions to Exercises
128
= –393.5 kJ + 2(–285.83 kJ) – (–74.8 kJ) – (0) = –890.4 kJ
4
4
4
4
2
CH
g
90
.
4
899
.
4
CH
mol
CH
g
04
.
16
kJ
4
.
890
CH
mol
1
kJ
10
720
.
2
=
=
×
−
×
×
5.101
(a)
Mg(s) + 2H
2
O(l)
→
Mg(OH)
2
(s) + H
2
(g)
)
s
(
Mg
H
)
l
(
O
H
H
2
(g)
H
H
)
s
(
Mg(OH)
H
H
f
2
f
2
f
2
f
rxn
o
o
You've reached the end of your free preview.
Want to read all 31 pages?
 Fall '07
 Wyzlouzil
 Thermodynamics, Enthalpy, Thermochemistry, Energy, Heat, ΔH