1 chapter 3 matter and energy 3.3temperature copyright © 2008 by pearson education, inc. publishing...
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Chapter 3 Matter and Energy
3.3Temperature
Copyright © 2008 by Pearson Education, Inc.Publishing as Benjamin Cummings
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Temperature
Temperature
• Is a measure of how hot or cold an object is compared to another object.
• 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
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Temperature Scales
Temperature scales • Are Fahrenheit,
Celsius, and Kelvin.
• Have reference points for the boiling and freezing points of water.
Copyright © 2008 by Pearson Education, Inc.Publishing as Benjamin Cummings
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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
Learning Check
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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
Solution
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• 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.
TF = 9/5 TC + 32
or TF = 1.8 TC + 32
Fahrenheit Formula
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Solving for °F Temperature
A person with hypothermia hasa 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
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• 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 TC
1.8 1.8
TF - 32 = TC
1.8
Celsius Formula
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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
Learning Check
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3) 41.0 °C
TC = (TF - 32°)
1.8
= (105.8° - 32°)
1.8
= 73.8° = 41.0°C
1.8
Solution
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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
Learning Check
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A pepperoni pizza is baked at 455°F. What
temperature is needed on the Celsius scale?
2) 235°C
TF - 32° = TC 1.8
(455° - 32°) = 235°C
1.8
Solution
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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
Learning Check
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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
Solution
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The kelvin temperatureHas 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 .
0 K = –273 °C
Kelvin Temperature Scale
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Temperatures
Table 3.6
Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings
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What is normal body temperature of 37°C in kelvins?
1) 236 K
2) 310. K
3) 342 K
Learning Check
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What is normal body temperature of 37°C in kelvins?
2) 310. K
TK = TC + 273
= 37°C + 273
= 310. K
Solution