Chapter 3 Matter and Energy
3.3
1
Temperature
Copyright © 2008 by Pearson Education, Inc.
Publishing as Benjamin Cummings
Temperature
¢ Is a measure of the hotness and
coldness of an object.
¢ It is a relative measure of the
thermal energy stored in a
substance.
¢ Indicates that heat flows from the
object with a higher temperature
to the object with a lower
temperature.
¢ Is measured using a
thermometer.
Copyright © 2008 by Pearson Education, Inc.
Publishing as Benjamin Cummings
2
Temperature Scales
Temperature scales
•  Kelvin used by
scientists.
•  Celsius used by
most of the world.
•  Fahrenheit used
by Americans
•  Have reference
points for the
boiling and
freezing points of
water.
3
Copyright © 2008 by Pearson Education, Inc.
Publishing as Benjamin Cummings
Learning Check
A. What is the temperature of freezing water?
1) 0°F
2) 0°C
3) 0 K
B. What is the temperature of boiling water?
1) 100°F
2) 32°F
3) 373 K
C. How many Celsius units are between the boiling
and freezing points of water?
1) 100
2) 180
3) 273
4
Solution
A. What is the temperature of freezing water?
2) 0°C
B. What is the temperature of boiling water?
3) 373 K
C. How many Celsius units are between the boiling
and freezing points of water?
1) 100
5
Fahrenheit Formula
•  On the Fahrenheit scale, there are 180°F
between the freezing and boiling points and
on the Celsius scale, there are 100°C.
180°F =
9°F =
1.8°F
100°C
5°C
1°C
•  In the formula for calculating the Fahrenheit
temperature, adding 32 adjusts the zero point
of water from 0°C to 32°F.
6
TF
= 9/5 TC + 32°
or TF
= 1.8 TC + 32°
Solving for °F Temperature
A person with hypothermia has
a body temperature of 34.8°C.
What is that temperature in °F?
TF = 1.8 TC
+ 32°
TF = 1.8 (34.8°C) + 32°
exact tenth's
exact
= 62.6 + 32°
= 94.6°F
tenth’s
Copyright © 2008 by Pearson Education, Inc.
Publishing as Benjamin Cummings
7
Celsius Formula
•  TC is obtained by rearranging the equation for TF.
TF
=
1.8TC + 32
•  Subtract 32 from both sides.
TF - 32
=
1.8TC ( +32 - 32)
TF - 32
=
1.8TC
•  Divide by 1.8 = °F - 32
1.8
TF - 32
1.8
8
=
TC
= 1.8 TC
1.8
Learning Check
The normal temperature of a chickadee is 105.8°F.
What is that temperature on the Celsius scale?
1) 73.8 °C
2) 58.8 °C
3) 41.0 °C
9
Solution
3) 41.0 °C
TC
=
=
=
10
(TF - 32°)
1.8
(105.8° - 32°)
1.8
73.8° = 41.0°C
1.8
Learning Check
A pepperoni pizza is baked at 455°F. What
temperature is needed on the Celsius scale?
1) 423°C
2) 235°C
3) 221°C
11
Solution
A pepperoni pizza is baked at 455°F. What
temperature is needed on the Celsius scale?
2) 235°C
TF - 32°
1.8
= TC
(455° - 32°) = 235°C
1.8
12
Learning Check
On a cold winter day, the temperature is –15°C.
What is that temperature in °F?
1) 19 °F
2) 59°F
3) 5°F
13
Solution
3) 5°F
TF = 1.8 TC + 32°
TF = 1.8(–15°C) + 32°
= – 27° + 32°
= 5°F
Note: Be sure to use the change sign key on
your calculator to enter the minus – sign.
1.8 x 15 +/ – = –27
14
Kelvin Temperature Scale
The kelvin temperature
Has 100 units between freezing and boiling points.
100 K = 100°C or
1 K = 1 °C
•  Adds 273 to the Celsius temperature.
TK
= TC + 273
•  Of 0 K (absolute zero) is the lowest possible
temperature .
0K
= –273 °C
15
Temperatures
Table 3.6
Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings
16
Learning Check
What is normal body temperature of 37°C in kelvins?
1) 236 K
2) 310. K
3) 342 K
17
Solution
What is normal body temperature of 37°C in kelvins?
2) 310. K
TK =
=
=
18
TC + 273
37°C + 273
310. K