18.1
HEAT, WORK, AND THE FIRST LAW
OF THERMODYNAMICS
E
XERCISES
Section 18.1 The First Law of Thermodynamics
15.
I
NTERPRET
We identify the system as the water in the insulated container. The problem is about work done to
raise the temperature of a system. The first law of thermodynamics is involved.
D
EVELOP
Since the container is perfectly insulated thermally, no heat enters or leaves the water in it. Thus,
Q
=
0
and the first law of thermodynamics in Equation 18.1 gives
∆
U
Q
W
W
=
−
= −
.
The change in the internal
energy of the water is determined from its temperature rise,
∆
∆
U
mc
T
=
(see comments in Section 16.1 on
internal energy).
E
VALUATE
The work done on the water is
W
U
mc
T
= −
= −
= −
⋅
= −
∆
∆
kg)(4.184 kJ/kg K)(7 K)
(1
29 3
.
kJ
A
SSESS
The negative sign signifies that work was done on the water.
16.
(a)
The change in the internal energy of the water is
∆
∆
U
mc T
=
,
and the work done by it (i.e., the negative of the
work done on it) is given. Therefore, Equation 18.1 gives
Q
U
W
=
+
=
⋅
+
∆
( .
0 5 kg)(4.184 kJ/kg K)(3 K)
(
.
.
.
−
=
−
= −
9 0
9 0
2 72
kJ)
6.28 kJ
kJ
kJ.
(The negative sign signifies that the water lost heat to its surroundings.)
(b)
If the water had been in perfect thermal isolation, no heat would have been transferred,
Q
=
0
and
W
U
= −
= −
∆
6 28
.
kJ
instead of
−
9 0
.
kJ.
17.
I
NTERPRET
We identify the system as the gas that undergoes expansion. The problem is about the change of
internal energy of a system and involves the first law of thermodynamics.
D
EVELOP
The heat added to the gas is
Q
Pt
=
=
=
(40
25
1000
W)(
s)
J.
In addition, the amount of work it does on
its surrounding is
W
=
750 J.
The change in internal energy can be found by using the first law of thermodynamics
given in Equation 18.1.
E
VALUATE
Using Equation 18.1 we find
∆
U
Q
W
=
−
=
−
=
1000
750
250
J
J
J
A
SSESS
Since
∆
U
>
0,
we conclude that the internal energy has increased.
18.
From Equation 18.2,
dQ dt
dU dt
dW dt
/
/
/
=
+
=
+
=
45 W
165 W
210 W.
19.
I
NTERPRET
This problem is about heat and mechanical energy, which are related by the first law of
thermodynamics. The system is the automobile engine.
D
EVELOP
Since we are dealing with rates, we make use of Equation 18.2:
dU
dt
dQ
dt
dW
dt
=
−
If we assume that the engine system operates in a cycle, then
dU dt
/
.
=
0
The engine's mechanical power output can
then be calculated once the heat output is known.
18
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18.2
Chapter 18
E
VALUATE
The above conditions yield
(
/
)
dQ dt
out
kW
=
68
and
(
)
0.17(
)
in
dW dt
dQ dt
/
/
=
.
Equation 18.2 then gives
dW
dt
dQ
dt
dQ
dt
dQ
dt
d
in
out
=
=
−
=
1
0 17
.
W
dt
dQ
dt
out
−
or
dW
dt
dQ dt
=
−
=
−
=
−
−
(
/
)
( .
( .
out
)
kW
)
0 17
1
68
0 17
1
1
1
1
3 9
.
kW
A
SSESS
We find the mechanical power output
dW dt
/
to be proportional to the heat output,
(
/
)
.
dQ dt
out
In addition,
dW dt
/
also increases with the percentage of the total energy released in burning gasoline that ends up as
mechanical work.
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 Fall '10
 Pheong
 Thermodynamics, Work, Adiabatic process

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