1. General, Organic, and Biological ChemistryCopyright © 2010 Pearson Education, Inc.1 Energy and Matter Temperature

2. General, Organic, and Biological ChemistryCopyright © 2010 Pearson Education, Inc.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

3. General, Organic, and Biological ChemistryCopyright © 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

4. General, Organic, and Biological ChemistryCopyright © 2010 Pearson Education, Inc.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. General, Organic, and Biological ChemistryCopyright © 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

6. General, Organic, and Biological ChemistryCopyright © 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. T F = 9 (T C ) + 32 ° 5 orT F = 1.8(T C ) + 32 ° Fahrenheit Formula

7. General, Organic, and Biological ChemistryCopyright © 2010 Pearson Education, Inc.7  T C is obtained by rearranging the equation for T F. T F = 1.8(T C ) + 32 °  Subtract 32 from both sides. T F – 32 ° = 1.8(T C ) + (32 ° – 32 °) T F – 32 ° = 1.8(T C )  Divide by 1.8. T F – 32 ° = 1.8 T C 1.8 1.8 T F – 32 ° = T C 1.8 Celsius Formula

8. General, Organic, and Biological ChemistryCopyright © 2010 Pearson Education, Inc.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

9. General, Organic, and Biological ChemistryCopyright © 2010 Pearson Education, Inc.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. General, Organic, and Biological ChemistryCopyright © 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 T C = T F – 32 ° 1.8 =(105.8 – 32 °) 1.8 =73.8 °F = 41.0 °C 1.8 ° tenth’s place Solution

11. General, Organic, and Biological ChemistryCopyright © 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

12. General, Organic, and Biological ChemistryCopyright © 2010 Pearson Education, Inc.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 one’s place Solution

13. General, Organic, and Biological ChemistryCopyright © 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

14. General, Organic, and Biological ChemistryCopyright © 2010 Pearson Education, Inc.14 3) 5 °F T F = 1.8T C + 32 ° T F = 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

15. General, Organic, and Biological ChemistryCopyright © 2010 Pearson Education, Inc.15 The Kelvin temperature  is obtained by adding 273 to the Celsius temperature T K = T C + 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

16. General, Organic, and Biological ChemistryCopyright © 2010 Pearson Education, Inc.16 Temperatures

17. General, Organic, and Biological ChemistryCopyright © 2010 Pearson Education, Inc.17 What is normal body temperature of 37 °C in kelvins? 1) 236 K 2) 310 K 3) 342 K Learning Check

18. General, Organic, and Biological ChemistryCopyright © 2010 Pearson Education, Inc.18 What is normal body temperature of 37 °C in kelvins? 2) 310 K T K = T C + 273 =37 °C + 273 =310. K one’s place Solution