•
Calculate the kinetic energy of a tennis ball traveling at 150 mph.
•
Assume that energy is converted entirely to heat, and compare to the energy required to increase the
temperature of water by 30°C.
The mass of a tennis ball is about 60 g or 6
×
10
–2
kg. Converting 150 mph to 68 m/s, we write
2
2
–2
2
2
2
2
1
1
kg m
KE
(6
10
kg)(68 m/s)
1.4
10
or 1.4
10
J
2
2
s
⋅
=
=
×
=
×
×
mu

CHAPTER 6:
THERMOCHEMISTRY
225
The energy required to heat 1 mL of water, which is 1 g of water, by 30°C is given by
q
=
ms t
= (1 g)(4.184 J/g·°C)(30°C) = 1.3
×
10
2
J
The answer is yes, barely.
6.151
Basic approach:
•
Look up the total volume of ocean water, or use the value we estimated in Problem 1.111.
•
Assume an average ocean temperature, take the boiling point of ocean water as 100°C, and calculate the
energy required to heat all of the ocean water to 100°C.
•
Look up the energy released by the sun in one second and compare to the value calculated to heat all of the
ocean water to 100°C.
We need to do a couple of web searches here. First, the total volume of ocean water is 1.3 billion km
3
or
3
9
3
21
3
1000 m
1000 L
1.3
10 km
1.3
10
L
1 km
1 m
×
×
×
=
×
(Alternatively, we could use the volume we calculated in Problem 1.111.)
The energy released by the sun in one second is about 4
×
10
26
J. Assume the average temperature of ocean
water to be 8°C (the surface ocean water is warm but deep down the ocean the water is quite cold) and the
specific heat of ocean water to be the same as that of water. The energy required to heat the water from 8°C
to 100°C is
q
=
ms t
= (1.3
×
10
21
L
×
1000 g/L)(4.184 J/g·
°
C)(100°C – 8°C) = 5
×
10
26
J
The answer is probably yes.
6.152
Basic approach:
•
An estimate of the volume of methane released is provided.
•
Solve for the moles of methane using the information provided.
•
Look up the enthalpy of combustion for methane and solve for the change in enthalpy.
Converting the temperature from
°
F to
°
C to K gives
5 C
(60
F
32
F)
16
C
(16
273)K
289 K
9 F
°
°
−
°
×
=
°
=
+
=
°
Converting cubic feet to liters gives
3
3
12
3
13
3
12 in
2.54 cm
1 mL
1 L
3
10
ft
8
10
L
1 ft
1 in
1000 mL
1 cm
×
×
×
×
×
=
×
Rearranging the ideal gas equation,
PV
=
nRT
, gives
13
12
(1 atm)(8
10
L)
3
10
mol
(0.0821 L atm/K mol)(289 K)
×
=
=
=
×
⋅
⋅
PV
n
RT
Note that a standard cubic foot of gas is around one mole of gas assuming ideal behavior.
CHAPTER 6:
THERMOCHEMISTRY
226
The enthalpy of combustion of methane is
−
890.4 kJ/mol; therefore, the energy that could be obtained from
three trillion cubic feet of methane would be
12
15
890.4 kJ
3
10
mol
3
10
kJ
1 mol
×
×
=
×
6.153
Basic approach:
•
Approximate the mass percent carbon in wood chips by looking up the formula for cellulose.
•
Solve for the mass of ethanol assuming 85 percent of the carbon is converted to ethanol.
•
Look up the enthalpy of combustion for ethanol and solve for enthalpy.

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- Thermodynamics, Enthalpy, Thermochemistry, ΔH, Standard enthalpy change of formation, kJ/mol, ΔH fD