Heat

•The big ideas!

Phases of Matter

• Solid – lattice work, has shape

• Liquid – fluid, takes shape of container

• Gas – fluid, needs sealed container

• (Plasma) – most energetic

Temperature

• Temperature – measure of hotnessdegrees Celsius or degrees Centigrade or °CFahrenheit scaleKelvin 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

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

Heat

• Heat is the flow of thermal energy from hot substance to cold substance related to the amount of thermal energyboth temperature and mass are important

• Objects in contact tend to reach same temperature thermal equilibrium

Heat

• Measure of energy transferred calorie – cal (little c)

Calorie = 1000 calories = kilocalorie (big C) Joule (1 calorie = 4.184 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

Specific Heat Capacity

• 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

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

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.

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 – 108,000 + 1,410Tf – 42,300 = 0

3,210Tf – 150,300 = 0

Tf = 150,300/3,210 = 46.8 C

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

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 – 486,000 + 470Tf – 14,100 = 0

8570Tf – 500100 = 0

Tf = 500,100/8,570 = 58.3 C

Heat Transfer• Conduction 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

Changing Phase

Gas

Liquid

SolidEnergy

meltingadd heat

freezingremove heat

evaporationadd heat

condensationremove heat

Heat of fusion80 cal/g for water334 kJ/kg

Heat of vaporization540 cal/g for water2257 kJ/kg

Freezing and Melting

Freezing 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

Evaporation and Condensation

• Evaporation 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

540 cal

80 cal

• 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?

• 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 2 g x 80 cal/g = 160 cal• 0 °C to 100 °C2g x 1 cal/g °C x 100 °C = 200 cal• Water to steam 2g x 540 cal/g = 1080 cal• total 1440 cal