1. Heat
The big ideas!

2. Phases of Matter Solid – lattice work, has shape
Liquid – fluid, takes shape of container
Gas – fluid, needs sealed container
(Plasma) – most energetic

3. Temperature Temperature – measure of hotness
degrees Celsius or degrees Centigrade or °C
Fahrenheit scale
Kelvin Scale (doesn’t use the  symbol)
Absolute zero
Freezing of water
Boiling of water
-273 °C
0 °C
100 °C
-460 °F
32 °F
212 °F
0 K
273 K
373 K

4. Temperature and Molecular Motion
The higher the temperature, the faster the molecules are moving.
The faster the molecules are moving, the more energy they have
For the same molecule: Gas > Liquid > Solid
Gas has more energy than a solid

5. Heat Heat is the flow of thermal energy
from hot substance to cold substance
related to the amount of thermal energy
both temperature and mass are important
Objects in contact tend to reach same temperature
thermal equilibrium

6. Heat Measure of energy transferred
calorie – cal (little c)
Calorie = 1000 calories = kilocalorie (big C)
Joule (1 calorie = J)
BTU (British Thermal Unit)
Transfer of Heat requires temperature difference
Heat moves from high to low temperature
Thermal expansion
most materials get bigger as they get hotter

7. Specific Heat Capacity
symbol “c”
Unit: Energy/mass * temperature difference
J/kg °C or cal/g °C or BTU/ lb °F
measures effect of heat on change in temperature
c of water is 1 cal/g °C this is high
c of metals is lower – easier to change temperature
Q = m c T

8. Thermal Equilibrium Problem
A 2.0 kg block of aluminum at 60 C is placed in contact with a 3.0 kg block of steel at 30 C. If no heat is lost to the environment, what will be the final temperature of the blocks?
cal = 900 J/kg C
csteel = 470 J/kg C

9. Thermal Equilibrium Problem
The aluminum block will lose heat to the steel block
The heat lost by the aluminum equals that gained by the steel
The final temperature will be some place between the initial temperatures of the blocks.

10. Qal + Qsteel = 0 (aluminum loses while steel gains)
mal cal Tal + mst cst Tst = 0
mal cal (Tf-Ti)al + mst cst (Tf-Ti)st = 0
2.0(900)(Tf-60) + 3.0(470)(Tf-30) =0
1,800Tf – 108, ,410Tf – 42,300 = 0
3,210Tf – 150,300 = 0
Tf = 150,300/3,210 = 46.8 C

11. Thermal Equilibrium Problem 2
A 9.0 kg block of aluminum at 60 C is placed in contact with a 1.0 kg block of steel at 30 C. If no heat is lost to the environment, what will be the final temperature of the blocks?
cal = 900 J/kg C
csteel = 470 J/kg C

12. Problem #2
Qal + Qsteel = 0 (aluminum loses while steel gains)
mal cal Tal + mst cst Tst = 0
mal cal (Tf-Ti)al + mst cst (Tf-Ti)st = 0
9.0(900)(Tf-60) + 1.0(470)(Tf-30) =0
8100Tf – 486, Tf – 14,100 = 0
8570Tf – = 0
Tf = 500,100/8,570 = 58.3 C

13. Heat Transfer Conduction Convection Radiation
Heat transfers along an object
metals have high conduction
Convection
heat transfers when a fluid moves
hot air rises and cold air sinks
creates wind and “weather”
Radiation
any object with thermal energy emits radiation
electromagnetic wave
black absorbs radiant energy better than white

14. Changing Phase Gas Liquid Solid Energy evaporation condensation
add heat
condensation
remove heat
Heat of vaporization
540 cal/g for water
2257 kJ/kg
Liquid
melting
add heat
Heat of fusion
80 cal/g for water
334 kJ/kg
freezing
remove heat
Solid

15. Freezing and Melting Freezing Melting temperature remains constant
from liquid to solid
loses energy (heat) to freeze a liquid
Melting
From solid to liquid
requires heat to melt solid
temperature remains constant
80 cal/g (334 kJ/kg) of heat lost to freeze water
Called Heat of fusion

16. Evaporation and Condensation
Heat required to change liquid to gas
boiling
Condensation
Heat released when gas change to liquid
temperature of solid and liquid are the same
540 cal/g (2257 kJ/kg) to evaporated water
Called Heat of vaporization

17. 540 cal
80 cal

18. How much heat is required to raise the temperature of 3 g of water from 23 °C to 39 ° C?
How much heat is required to melt 3 g of ice at 0 ° C?
How much heat is required to vaporize 4 g of water to steam at
100 ° C?
How much heat is required to take 2 g of ice from 0 ° C to steam at 100 ° C?

19. How much heat is required to raise the temperature of 3 g of water from 23 °C to 39 °C?
Q=mc(Tf-Ti)= 3g x 1 cal/g °C x 16 °C = 48 cal
How much heat is required to melt 3 g of ice at 0 °C?
3 g x 80 cal/g = 240 cal
How much heat is required to vaporize 4 g of water to steam at
100 °C?
4 g x 540 cal/g = 2160 cal
How much heat is required to take 2 g of ice from 0 °C to steam at 100 °C?
Ice to water g x 80 cal/g = 160 cal
0 °C to 100 °C 2g x 1 cal/g °C x 100 °C = 200 cal
Water to steam g x 540 cal/g = cal
total cal