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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):
9
5
( 15) 32
5 F.
F
T
=

+
= °
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:
9
32
5
F
C
T
T
=
+
E
VALUATE
Solving the above equation for the Celsius temperature, we obtain
5
5
(
32)
(68 32)
20 C
9
9
C
F
T
T
=

=

=
°
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
(9/5)
,
F
C
T
T
∆
=
∆
9
5
so ( )(10 C )
18 F.
° =
°
(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
F
T
and
C
T
are numerically equal.
D
EVELOP
The two temperature scales are related by Equation 16.2:
9
32
5
F
C
T
T
=
+
The condition that the readings are the same numerically is
9
5
( )
32
.
F
C
C
T
T
T
=
+
=
E
VALUATE
The above equation can be solved to give
5
(32)
40
4
C
F
T
T
= 
= 
=
A
SSESS
This is the only temperature in which both scales yield the same reading:
40 F
40 C.

° = 
°
18.
Equations 16.1 and 16.2 give
77.3 273.15
196 C,
C
T
=


°
.
and
9
5
( )( 196) 32
321 F.
F
T
=

+
= 
°
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:
9
5
32.
F
C
T
T
=
+
E
VALUATE
Solving the above equation for the Fahrenheit temperature, we obtain
9
(39.1) 32
102.4 F
5
F
T
=
+
=
°
A
SSESS
The temperature is way above the normal body temperature of
98.6 F
°
(or
37 C
°
). Call the doctor
immediately!
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.
16.1
16
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View Full Document 16.2
Chapter 16
D
EVELOP
The specific heat of concrete is given in Table 16.1 as
880 J/kg K.
c
=
⋅
We multiply this by the mass
55,000 kg
m
=
of the block to find the heat capacity of the block.
E
VALUATE
6
48 10 J/K.
C
mc
=
=
×
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
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This note was uploaded on 09/12/2009 for the course PH PH101 taught by Professor Prof.kim during the Spring '09 term at Korea Advanced Institute of Science and Technology.
 Spring '09
 Prof.Kim
 Physics, Heat

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