16.1
TEMPERATURE AND HEAT
E
XERCISES
Section 16.1 Heat, Temperature, and Thermodynamic Equilibrium
14.
We assume that the U.S. meteorologist predicts the same temperature, but expresses it on the Fahrenheit scale
(Equation 16.2):
T
F
=
−
+
=
°
9
5
15
32
5
(
)
.
F
15.
I
NTERPRET
This problem is about converting temperature from the Fahrenheit scale to the Celsius scale.
D
EVELOP
The two temperature scales are related by Equation 16.2:
T
T
F
C
=
+
9
5
32
E
VALUATE
Solving the above equation for the Celsius temperature, we obtain
T
T
C
F
=
−
=
−
=
°
5
9
32
5
9
68
32
20
(
)
(
)
C
A
SSESS
This is a useful result to remember, since
20
°
C,
or
68
°
F
is a typical room temperature.
16.
Temperature differences on the Fahrenheit and Celsius scales are related by
∆
∆
T
T
F
C
=
( / )
,
9 5
so
C
F .
( )(
)
9
5
10
18
° =
°
(Note that a temperature difference and a temperature reading are not the same, even though
both are specified in the same units. The notation
C
°
versus
°
C
is an attempt to clarify this distinction, but is not
universally accepted or consistently applied.)
17.
I
NTERPRET
Given both Fahrenheit and Celsius scales, we want to know when
T
F
and
T
C
are numerically equal.
D
EVELOP
The two temperature scales are related by Equation 16.2:
T
T
F
C
=
+
9
5
32
The condition that the readings are the same numerically is
T
T
T
F
C
C
=
+
=
( )
.
9
5
32
E
VALUATE
The above equation can be solved to give
T
T
C
F
= −
= −
=
5
4
32
40
(
)
A
SSESS
This is the only temperature in which both scales yield the same reading:
−
°
= −
°
40
40
F
C.
18.
Equations 16.1 and 16.2 give
T
C
=
−
−
°
77 3
273 15
196
.
.
C,
and
T
F
=
−
+
= −
°
( )(
)
9
5
196
32
321 F.
19.
I
NTERPRET
This problem is about converting temperature from the Celsius scale to the Fahrenheit scale.
D
EVELOP
The two temperature scales are related by Equation 16.2:
T
T
F
C
=
+
9
5
32.
E
VALUATE
Solving the above equation for the Fahrenheit temperature, we obtain
T
F
=
+
=
°
9
5
39 1
32
102 4
(
. )
.
F
A
SSESS
The temperature is way above the normal body temperature of
98 6
.
°
F
(or
37
°
C
). Call the doctor
immediately!
16
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16.2
Chapter 16
Section 16.2 Heat Capacity and Specific Heat
20.
I
NTERPRET
We find the heat capacity of a large concrete block. We know the mass of the block and its specific heat.
D
EVELOP
The specific heat of concrete is given in Table 16.1 as
c
=
⋅
880 J/kg K.
We multiply this by the mass
m
=
55 000
,
kg
of the block to find the heat capacity of the block.
E
VALUATE
C
mc
=
=
×
48
10
6
J/K.
A
SSESS
This is a large value, but then it takes a large amount of heat to change the temperature of a 55tonne
block of concrete.
21.
I
NTERPRET
We find the energy necessary to change the temperature of an object by a given amount. This
involves the heat capacity of the object and the temperature.
D
EVELOP
We use the equation
Q
mc T
=
∆
.
The mass of the aluminum block is
m
=
2 0
.
kg,
the specific heat is
c
=
⋅
900 J/kg K,
and the temperature change is
∆
T
=
18 C°.
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 Fall '10
 Pheong
 Thermodynamics, Energy, Heat

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