general, organic, and biological chemistrycopyright © 2010 pearson education, inc.1 energy and...
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General, Organic, and Biological Chemistry Copyright © 2010 Pearson Education, Inc. 1
Energy and Matter
Temperature
General, Organic, and Biological Chemistry Copyright © 2010 Pearson Education, Inc. 2
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
General, Organic, and Biological Chemistry Copyright © 2010 Pearson Education, Inc. 3
Temperature Scales
Temperature scales are Fahrenheit,
Celsius, and Kelvin have reference
points for the boiling and freezing points of water
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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
General, Organic, and Biological Chemistry Copyright © 2010 Pearson Education, Inc. 5
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
General, Organic, and Biological Chemistry Copyright © 2010 Pearson Education, Inc. 6
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 the Fahrenheit temperature, adding 32 adjusts the zero point of water from 0 °C to 32 °F.
TF = 9 (TC) + 32 ° 5
or TF = 1.8(TC) + 32 °
Fahrenheit Formula
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TC is obtained by rearranging the equation for TF.
TF = 1.8(TC) + 32 °
Subtract 32 from both sides.
TF – 32 ° = 1.8(TC) + (32 ° – 32 °)
TF – 32 ° = 1.8(TC)
Divide by 1.8. TF – 32 ° = 1.8 TC
1.8 1.8
TF – 32 ° = TC 1.8
Celsius Formula
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Solving a Temperature Problem
A person with hypothermia has abody temperature of 34.8 °C. WhatIs 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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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
General, Organic, and Biological Chemistry Copyright © 2010 Pearson Education, Inc. 10
The normal temperature of a chickadee is 105.8 °F.
What is that temperature on the Celsius scale?
3) 41.0 °C
TC = TF – 32 °
1.8
= (105.8 – 32 °)
1.8
= 73.8 °F = 41.0 °C
1.8 ° tenth’s place
Solution
General, Organic, and Biological Chemistry Copyright © 2010 Pearson Education, Inc. 11
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
General, Organic, and Biological Chemistry Copyright © 2010 Pearson Education, Inc. 12
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 one’s place
Solution
General, Organic, and Biological Chemistry Copyright © 2010 Pearson Education, Inc. 13
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 = 1.8TC + 32 °
TF = 1.8(–15 °C) + 32 °
= – 27 + 32 °
= 5 °F one’s place
Note: Be sure to use the change sign key on your calculator to enter the minus (–) sign.
1.8 x 15 +/ – = –27
Solution
General, Organic, and Biological Chemistry Copyright © 2010 Pearson Education, Inc. 15
The Kelvin temperature is obtained by adding 273 to the Celsius temperature
TK = TC + 273
In the Kelvin temperature scale: There are 100 units between the freezing and boiling
points of water.
100 K = 100 °C or 1 K = 1 °C 0 K (absolute zero) is the lowest possible
temperature.
0 K = –273 °C
Kelvin Temperature Scale
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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