1. Energy in Thermal Processes
Thermal Physics
Energy in Thermal Processes

2. Energy Transfer
When two objects of different temperatures are placed in thermal contact, the temperature of the warmer decreases and the temperature of the cooler increases
The energy exchange ceases when the objects reach thermal equilibrium
The concept of energy was broadened from just mechanical to include internal
Made Conservation of Energy a universal law of nature

3. Heat Compared to Internal Energy
Important to distinguish between them
They are not interchangeable
They mean very different things when used in physics

4. Internal Energy
Internal Energy, U, is the energy associated with the microscopic components of the system
Includes kinetic and potential energy associated with the random translational, rotational and vibrational motion of the atoms or molecules
Also includes any potential energy bonding the particles together

5. Thermal Energy
Thermal Energy is the portion of the Internal Energy, U, that is associated with the motion of the microscopic components of the system.

6. Heat
Heat is the transfer of energy between a system and its environment because of a temperature difference between them
The symbol Q is used to represent the amount of energy transferred by heat between a system and its environment

7. Units of Heat
Calorie
An historical unit, before the connection between thermodynamics and mechanics was recognized
A calorie is the amount of energy necessary to raise the temperature of 1 g of water from 14.5° C to 15.5° C .
A Calorie (food calorie) is 1000 cal

8. Units of Heat, cont. US Customary Unit – BTU
BTU stands for British Thermal Unit
A BTU is the amount of energy necessary to raise the temperature of 1 lb of water from 63° F to 64° F
1 cal = J
This is called the Mechanical Equivalent of Heat

9. Problem: Working Off Breakfast
A student eats breakfast consisting of two bowls of cereal and milk, containing a total of 3.20 x 102 Calories of energy. He wishes to do an equivalent amount of work in the gymnasium by doing curls with a 25 kg barbell. How many times must he raise the weight to expend that much energy? Assume that he raises it through a vertical displacement of 0.4 m each time, the distance from his lap to his upper chest. h

10. Problem: Working Off Breakfast
Convert his breakfast Calories, E, to joules:

11. Problem: Working Off Breakfast
Use the work-energy theorem to find the work necessary to lift the barbell up to its maximum height.
The student must expend the same amount of energy lowering the barbell, making 2mgh per repetition. Multiply this amount by n repetitions and set it equal to the food energy E:

12. Problem: Working Off Breakfast
Solve for n, substituting the food energy for E:

13. James Prescott Joule 1818 – 1889 British physicist
Conservation of Energy
Relationship between heat and other forms of energy transfer

14. Specific Heat
Every substance requires a unique amount of energy per unit mass to change the temperature of that substance by 1° C
The specific heat, c, of a substance is a measure of this amount

15. Units of Specific Heat
SI units
J / kg °C
Historical units
cal / g °C

16. Heat and Specific Heat Q = m c ΔT
ΔT is always the final temperature minus the initial temperature
When the temperature increases, ΔT and ΔQ are considered to be positive and energy flows into the system
When the temperature decreases, ΔT and ΔQ are considered to be negative and energy flows out of the system

17. A Consequence of Different Specific Heats
Water has a high specific heat compared to land
On a hot day, the air above the land warms faster
The warmer air flows upward and cooler air moves toward the beach

18. Calorimeter
One technique for determining the specific heat of a substance
A calorimeter is a vessel that is a good insulator which allows a thermal equilibrium to be achieved between substances without any energy loss to the environment

19. Calorimetry Analysis performed using a calorimeter
Conservation of energy applies to the isolated system
The energy that leaves the warmer substance equals the energy that enters the water
Qcold = -Qhot
Negative sign keeps consistency in the sign convention of ΔT

20. Calorimetry with More Than Two Materials
In some cases it may be difficult to determine which materials gain heat and which materials lose heat
You can start with SQ = 0
Each Q = m c DT
Use Tf – Ti
You don’t have to determine before using the equation which materials will gain or lose heat

21. Phase Changes
A phase change occurs when the physical characteristics of the substance change from one form to another
Common phases changes are
Solid to liquid – melting
Liquid to gas – boiling
Phases changes involve a change in the internal energy, but no change in temperature

22. Latent Heat During a phase change, the amount of heat is given as
Q = ±m L
L is the latent heat of the substance
Latent means hidden
L depends on the substance and the nature of the phase change
Choose a positive sign if you are adding energy to the system and a negative sign if energy is being removed from the system

23. Latent Heat, cont. SI units of latent heat are J / kg
Latent heat of fusion, Lf, is used for melting or freezing
Latent heat of vaporization, Lv, is used for boiling or condensing
Table 11.2 gives the latent heats for various substances

24. Problem: Boiling Liquid Helium
Liquid helium has a very low boiling point, 4.2 K, as well as low latent heat of vaporization, 2.09 x 104 J/kg. If energy is transferred to a container of liquid helium at the boiling point from an immersed electric heater at a rate of 10 W, how long does it take to boil away 2 kg of the liquid?

25. Problem: Boiling Liquid Helium
Find the energy needed to vaporize 2 kg of liquid helium at its boiling point:
Divide this result by the power to find the time:

26. Sublimation
Some substances will go directly from solid to gaseous phase
Without passing through the liquid phase
This process is called sublimation
There will be a latent heat of sublimation associated with this phase change

27. Graph of Ice to Steam

28. Warming Ice Start with one gram of ice at –30.0º C
During A, the temperature of the ice changes from –30.0º C to 0º C
Use Q = m c ΔT
Will add 62.7 J of energy

29. Melting Ice Once at 0º C, the phase change (melting) starts
The temperature stays the same although energy is still being added
Use Q = m Lf
Needs 333 J of energy

30. Warming Water
Between 0º C and 100º C, the material is liquid and no phase changes take place
Energy added increases the temperature
Use Q = m c ΔT
419 J of energy are added

31. Boiling Water At 100º C, a phase change occurs (boiling)
Temperature does not change
Use Q = m Lv
2 260 J of energy are needed

32. Heating Steam
After all the water is converted to steam, the steam will heat up
No phase change occurs
The added energy goes to increasing the temperature
Use Q = m c ΔT
To raise the temperature of the steam to 120°, 40.2 J of energy are needed

33. Problem Solving Strategies
Make a table
A column for each quantity
A row for each phase and/or phase change
Use a final column for the combination of quantities
Use consistent units

34. Problem Solving Strategies, cont
Apply Conservation of Energy
Transfers in energy are given as Q=mcΔT for processes with no phase changes
Use Q = m Lf or Q = m Lv if there is a phase change
In Qcold = - Qhot be careful of sign
ΔT is Tf – Ti
Solve for the unknown

35. Your Turn
You start with 250. g of ice at -10 C. How much heat is needed to raise the temperature to 0 C?
10.5 kJ
How much more heat would be needed to melt it?
83.3 kJ

36. Your Turn
You start with 250. g of ice at -10 C. What will happen if we add 50. kJ of heat?
10.5 kJ will be used to warm it up to the MP, and the rest will start melting the ice.
0.119 kg will be melted

37. Problem: Partial melting
A 5 kg block of ice at 0o C is added to an insulated container partially filled with 10 kg of water at 15 o C.
(a) Find the temperature, neglecting the heat capacity of the container.
(b) Find the mass of the ice that was melted.

38. Problem: Partial melting
(a) Find the equilibrium temperature.
First, Compute the amount of energy necessary to completely melt the ice:

39. Problem: Partial melting
Next, calculate the maximum energy that can be lost by the initial mass of liquid water without freezing it:
This is less than half the energy necessary to melt all the ice, so the final state of the system is a mixture of water and ice at the freezing point:

40. Problem: Partial melting
(b) Compute the mass of the ice melted.
Set the total available energy equal to the heat of fusion of m grams of ice, mLf:

41. Final Problem
100. grams of hot water ( 60. C) is added to a 1.0 kg iron skillet at 500 C. What is the final temperature and state of the mixture?

42. Final Problem 16.7 kJ needed to warm water to BP.
226 kJ needed to vaporize water
199.2 kJ will be given up by skillet.
Final temperature will be 100. C
182 kJ of heat from the skillet will be available to vaporize water
81 grams of water will vaporize.