Chapter 2 Energy and Matter 2.2 Temperature

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

Temperature Scales Temperature scales are Fahrenheit, Celsius, and Kelvin have reference points for the boiling and freezing points of water

Learning Check 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

Solution 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

Fahrenheit Formula 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 °

Celsius Formula 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

Solving a Temperature Problem A person with hypothermia has a 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

Learning Check 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

Solution 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 °) = 73.8 °F = 41.0 °C 1.8 ° tenth’s place

Learning Check 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

Solution 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

Learning Check 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

Solution 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

Kelvin Temperature Scale 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

Temperatures

Learning Check What is normal body temperature of 37 °C in kelvins?

Solution What is normal body temperature of 37 °C in kelvins? 2) 310 K TK = TC + 273 = 37 °C + 273 = 310. K one’s place