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

2. General, Organic, and Biological ChemistryCopyright © 2010 Pearson Education, Inc. Temperature Temperature is a measure of the average heat or thermal energy of the particles in a substance. Since it is an average measurement, it does not depend on the number of particles in an object. In that sense it does not depend on the size of it. For example, the temperature of a small cup of boiling water is the same as the temperature of a large pot of boiling water. Even if the large pot is much bigger than the cup and has millions and millions more water molecules. 2

3. General, Organic, and Biological ChemistryCopyright © 2010 Pearson Education, Inc.3 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

4. General, Organic, and Biological ChemistryCopyright © 2010 Pearson Education, Inc.4 Temperature Scales Temperature scales  are Fahrenheit, Celsius, and Kelvin  have reference points for the boiling and freezing points of water

5. General, Organic, and Biological ChemistryCopyright © 2010 Pearson Education, Inc.5 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

6. General, Organic, and Biological ChemistryCopyright © 2010 Pearson Education, Inc.6 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

7. General, Organic, and Biological ChemistryCopyright © 2010 Pearson Education, Inc.7  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

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

9. General, Organic, and Biological ChemistryCopyright © 2010 Pearson Education, Inc.9 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

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? 1) 73.8 °C 2) 58.8 °C 3) 41.0 °C Learning Check

11. General, Organic, and Biological ChemistryCopyright © 2010 Pearson Education, Inc.11 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

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? 1) 423 °C 2) 235 °C 3) 221 °C Learning Check

13. General, Organic, and Biological ChemistryCopyright © 2010 Pearson Education, Inc.13 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

14. General, Organic, and Biological ChemistryCopyright © 2010 Pearson Education, Inc.14 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

15. General, Organic, and Biological ChemistryCopyright © 2010 Pearson Education, Inc.15 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

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

17. General, Organic, and Biological ChemistryCopyright © 2010 Pearson Education, Inc.17 Temperatures

18. General, Organic, and Biological ChemistryCopyright © 2010 Pearson Education, Inc.18 What is normal body temperature of 37 °C in kelvins? 1) 236 K 2) 310 K 3) 342 K Learning Check

19. General, Organic, and Biological ChemistryCopyright © 2010 Pearson Education, Inc.19 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