1. Thermal Energy

2. Thermal Energy Temperature & Heat
Temperature is related to the average kinetic energy of the particles in a substance.

3. SI unit for temp. is the Kelvin
K = °C (10°C = 283K)
°C = K – 273 (10K = -263°C)
Thermal Energy – the total of all the kinetic and potential energy of all the particles in a substance.

4. Thermal energy relationships
As temperature increases, so does thermal energy (because the kinetic energy of the particles increased).
Even if the temperature doesn’t change, the thermal energy in a more massive substance is higher (because it is a total measure of energy).

5. Heat The flow of thermal energy from one object to another.
Cup gets cooler while hand gets warmer
Heat
The flow of thermal energy from one object to another.
Heat always flows from warmer to cooler objects.
Ice gets warmer while hand gets cooler

6. Types of Heat Transfer

7. 3 WAYS THAT HEAT CAN TRAVEL SONG:

8. Types of Heat Transfer Conduction
Conduction is the transfer of heat by direct contact. *occurs best in solids

9. Types of Heat Transfer Convection
Convection is the transfer of heat through fluids by currents created due to density differences in parts of the fluid. *occurs in liquids and gases only
Hot water rises, cools and falls.
Heated air rises, cools and then falls.
Cool air falls.

10. Types of Heat Transfer Radiation
Radiation is the transfer of heat energy through space as waves of electromagnetic radiation. *the only type that can occur through empty space (example: from the sun)

11. Land heats up and cools down faster than water.
Specific Heat
Some things heat up or cool down faster than others.
Land heats up and cools down faster than water.

12. Specific heat is the amount of heat required to raise the temperature of 1 kg of a material by one degree (C or K).
C water = 4184 J / kg C
C sand = 664 J / kg C
This is why land heats up quickly during the day and cools quickly at night and why water takes longer.

13. Specific Heat 50g Al 50g Cu Which take longer to heat to 100°C?
(cal/g°C)
(J/g°C)
Water
1.00
4.18
Aluminum
0.22
0.90
Copper
0.093
0.39
Silver
0.057
0.24
Gold
0.031
0.13
50g Al 50g Cu
Aluminum has a higher specific heat, so it will take longer to heat up. It will ALSO take longer to cool down.

14. Why does water have such a high specific heat?
water metal
Water molecules form strong bonds with each other; therefore it takes more heat energy to break them. Metals have weak bonds and do not need as much energy to break them.

15. -q means heat loss +q = heat gain
How to calculate changes in thermal energy
q = mCpT
q = change in thermal energy (heat)
m = mass of substance
T = change in temperature (Tf – Ti)
Cp = specific heat of substance
-q means heat loss +q = heat gain

16. Heat Transfer A 32g silver spoon cools from 60°C to 20°C
Heat Transfer A 32g silver spoon cools from 60°C to 20°C. How much heat is lost by the spoon?
GIVEN
m = 32g
Ti = 60°C
Tf = 20°C
q = ??
Cp = J/g°C
WORK
q = mCpΔT
m = 32g
ΔT = Tf - Ti
ΔT = 20°C – 60°C = -40°C
q = (32g)(0.235J/g°C)(-40°C)
q = J

17. Heat Transfer How much heat is required to warm 230 g of water from 12°C to 90°C?
GIVEN
m = 230g
Ti = 12°C
Tf = 90°C
Q = ??
Cp = J/g°C
WORK
q = mCpΔT
m = 230g
ΔT = Tf - Ti
ΔT = 90°C – 12°C = 78°C
q = (230g)(4.184J/g°C)(78°C)
q = 75,061 J

18. A piece of iron at a temperature of 145°C cools off to 45°C
A piece of iron at a temperature of 145°C cools off to 45°C. If the iron has a mass of 10 g and a specific heat of J/g°C, how much heat is given up?
GIVEN
m = 10g
Ti = 145°C
Tf = 45°C
q = ??
Cp = J/g°C
WORK
q = mCpΔT
m = 10g
ΔT = Tf - Ti
ΔT = 45°C – 145°C = -100°C
q = (10g)(0.449J/g°C)(-100°C)
q = -449 J

19. First, mass and temperature of water are measured
A calorimeter is used to help measure the specific heat of a substance.
Heat gained = Heat lost
First, mass and temperature of water are measured
Knowing its q value, its mass, and its T, its Cp can be calculated
Then heated sample is put inside and heat flows into water
This gives the heat lost by the substance
T is measured for water to help get its heat gain

20. Let’s Practice!
A 55.1 g piece of metal is heated to a temp of 45.1°C, and placed into a cup containing 359g of water at 20.0°C. The final temp of the water and metal is 22.3°C.
How much heat energy did the water absorb?
q = mcΔT
q = (359g)(4.18J/g°C)(22.3°C – 20.0°C) = 3.45 x 103J
How much heat energy did the metal release to the water?
q lost = q gained
q lost by the metal = x 103J
The q is negative because heat was lost.
What is the specific heat of the metal?
3.45 x 103J = (55.1g)(C)(22.3°C – 45.1°C)
2.75 J/g°C = C

21. Potential Energy Diagrams

22. Endothermic Reactions
The reactants have less potential energy than the products do. Energy must be input in order to raise the particles up to the higher energy level.
Energy + A + B --> AB

24. Exothermic Reactions
The reactants have more potential energy than the products have. The extra energy is released to the surroundings.
A + B --> AB + Energy

26. Heat of Fusion
The Quantity of heat absorbed when a specific quantity of the solid is converted to liquid at its melting point is called its heat of fusion.

27. Heat of Vaporization
The quantity of heat absorbed when a specific quantity of the liquid is converted to gas at the boiling point is called the heat of vaporization.

28. Phase Diagram Gas Liquid Solid Water boils Ice melts Vaporization
Fusion

29. Hess's law
The energy change in an overall chemical reaction is equal to the sum of the energy changes in the individual reactions comprising it.

30. To Solve Hess’s Law Problems
Write the balanced chemical equation.
Look up the heat of formation in a table.
Substances in their elemental form are always zero (EX. O2, C, H2, S all have a ∆H of 0)
Multiply the heat of formation by the number of moles.
Add the products together.
Add the reactants together.
Subtract the products from the reactants.

31. Heat of Formation Table

32. Example 1
For the reaction of oxygen and carbon yielding carbon dioxide, balance and then calculate ΔH and classify as endothermic or exothermic:
__O2 + __C ⇌ __CO2
When we plug in our equation for the formation of CO2:
ΔHreaction= ΔHfo[CO2] - (ΔHfo[O2] + ΔHfo[C])
(0 + 0)
ΔHfo[CO2]=-393.5kJ exothermic

33. Example 2 _2_NO + __O2 ⇌ __NO2
For the reaction of oxygen and nitrogen monoxide yielding nitrogen dioxide, balance and then calculate ΔH and classify as endothermic or exothermic:
_2_NO + __O2 ⇌ __NO2
When we plug in our equation for the formation of NO2:
ΔHreaction= ΔHfo[NO2] - (ΔHfo[O2] + ΔHfo[NO])
(33.18) - ( *2)
ΔHfo[NO2]= kJ exothermic