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1 Chapter 1 Matter, Measurements, & Calculations 1.6 cont’ Temperature Conversions Copyright © 2005 by Pearson Education, Inc. Publishing as Benjamin Cummings.

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Presentation on theme: "1 Chapter 1 Matter, Measurements, & Calculations 1.6 cont’ Temperature Conversions Copyright © 2005 by Pearson Education, Inc. Publishing as Benjamin Cummings."— Presentation transcript:

1 1 Chapter 1 Matter, Measurements, & Calculations 1.6 cont’ Temperature Conversions Copyright © 2005 by Pearson Education, Inc. Publishing as Benjamin Cummings

2 2 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 © 2005 by Pearson Education, Inc. Publishing as Benjamin Cummings

3 3 Temperature Scales Copyright © 2005 by Pearson Education, Inc. Publishing as Benjamin Cummings are Fahrenheit, Celsius, and Kelvin. have reference points for the boiling and freezing points of water.

4 4 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) 1002) 1803) 273 Learning Check

5 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

6 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. T F = 9/5 T C + 32  or T F = 1.8 T C + 32  Fahrenheit Formula

7 7 T C is obtained by rearranging the equation for T F. T F = 1.8T C + 32 Subtract 32 from both sides. T F - 32 = 1.8T C ( +32 - 32) T F - 32 = 1.8T C Divide by 1.8 =°F - 32 = 1.8 T C 1.8 1.8 T F - 32 = T C 1.8 Celsius Formula

8 8 Solving A Temperature Problem A person with hypothermia has a body temperature of 34.8°C. What is that temperature in °F? T F = 1.8 T C + 32  T F = 1.8 (34.8°C) + 32° exact tenth's exact = 62.6 + 32° = 94.6°F tenth’s Copyright © 2005 by Pearson Education, Inc. Publishing as Benjamin Cummings

9 9 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

10 10 3) 41.0 °C T C = (T F - 32°) 1.8 =(105.8 - 32°) 1.8 =73.8°F = 41.0°C 1.8° Solution

11 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

12 12 A pepperoni pizza is baked at 455°F. What temperature is needed on the Celsius scale? 2) 235°C T F - 32° = T C 1.8 (455 - 32°) = 235°C 1.8 Solution

13 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

14 14 3) 5°F T F = 1.8 T C + 32  T F = 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

15 15 The Kelvin temperature scale has 100 units between the freezing and boiling points of water. 100 K = 100°Cor 1 K = 1°C is obtained by adding 273 to the Celsius temperature. T K = T C + 273 contains the lowest possible temperature, absolute zero (0 K). 0 K = –273°C Kelvin Temperature Scale

16 16 Temperatures TABLE 2.5

17 17 What is normal body temperature of 37°C in Kelvins? 1) 236 K 2) 310. K 3)342 K Learning Check

18 18 What is normal body temperature of 37°C in kelvins? 2) 310. K T K = T C + 273 = 37°C + 273 = 310. K Solution


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