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16.1
TEMPERATURE AND HEAT
EXERCISES
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.
INTERPRET
This problem is about converting temperature from the Fahrenheit scale to the Celsius scale.
DEVELOP
The two temperature scales are related by Equation 16.2:
TT
FC
=+
9
5
32
EVALUATE
Solving the above equation for the Celsius temperature, we obtain
CF
=−
=
°
5
9
32
5
9
68
32
20
C
ASSESS
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
∆∆
=
(/)
,
95
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.
INTERPRET
Given both Fahrenheit and Celsius scales, we want to know when
T
F
and
T
C
are numerically equal.
DEVELOP
The two temperature scales are related by Equation 16.2:
9
5
32
The condition that the readings are the same numerically is
T
C
=
.
9
5
32
EVALUATE
The above equation can be solved to give
=
5
4
32
40
ASSESS
This is the only temperature in which both scales yield the same reading:
−°=
−°
40
40
.
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.
INTERPRET
This problem is about converting temperature from the Celsius scale to the Fahrenheit scale.
DEVELOP
The two temperature scales are related by Equation 16.2:
9
5
32.
EVALUATE
Solving the above equation for the Fahrenheit temperature, we obtain
T
F
=
°
9
5
39 1
32
102 4
(.
)
.F
ASSESS
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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View Full Document 16.2
Chapter 16
Section 16.2 Heat Capacity and Specific Heat
20.
INTERPRET
We find the heat capacity of a large concrete block. We know the mass of the block and its specific heat.
DEVELOP
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
,k
g
of the block to find the heat capacity of the block.
EVALUATE
Cm
c
==
×
48
10
6
J/K.
ASSESS
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.
INTERPRET
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.
DEVELOP
We use the equation
Qm
cT
=∆
. The mass of the aluminum block is
m
=
20
. kg, the specific heat is
c
900 J/kg K, and the temperature change is
∆
T
=
18 C°.
EVALUATE
∆
32 000
,J
.
ASSESS
The same value would be the heat released by the aluminum if it cooled 18 C°.
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This note was uploaded on 10/21/2010 for the course PHYSICS 2131441 taught by Professor Pheong during the Fall '10 term at University of California, Berkeley.
 Fall '10
 Pheong

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