1. 1.7 Temperature

2. 373.15 212 100 H2O b.p. 100 -0 --------- 100 divisions 212 -32 ------------ 180 divisions 373.15 -273.15 ------------ 100 divisions 0 273.15 32 H2O f.p. F Fahrenheit C Celsius K Kelvin

3. Temperature Changing from one exact temperature to another C and K Each scale has 100 divisions (marks) BUT we need to adjust for the different zero point Therefore, we add/subtract 273.15 K= C + 273.15 C= K – 273.15 F and C Divisions and zero point are different so TWO adjustments are needed

4. F and C Adjustments 180 F/100 C 212 180 100 180 divisions (F) 1.8 F/1 C -32 A Conversion Factor 0 32 0 +32 Use conversion factor Adjust for 0 point C F

5. Examples 70.0 F= __________ K 70.0 F - 32 F=38.0 F x 1C/1.8 F= 21.1C +273.15 294.3 K A change in temperature RANGE (*just use conversions*) Note: Δ = “a change in”

6. Examples Cont’d A change in 20.0 Kelvin is a change in how many C? 20.0 K 1C 20.0C 1 K

7. Examples Cont’d A change in 30.0C is a change in how many F? 1.8F 30.0C 54.0F 1C