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## PowerPoint Slideshow about ' Thermal Physics' - kelsey-preston

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### Thermal Physics

Energy in Thermal Processes

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

Heat Compared to Internal Energy

- Important to distinguish between them
- They are not interchangeable

- They mean very different things when used in physics

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

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.

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

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

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 = 4.186 J
- This is called the Mechanical Equivalent of Heat

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

Problem: Working Off Breakfast

- Convert his breakfast Calories, E, to joules:

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:

Problem: Working Off Breakfast

- Solve for n,substituting the food energy for E:

James Prescott Joule

- 1818 – 1889
- British physicist
- Conservation of Energy
- Relationship between heat and other forms of energy transfer

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

Units of Specific Heat

- SI units
- J / kg °C

- Historical units
- cal / g °C

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

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

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

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

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

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

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

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

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?

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:

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

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

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

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

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

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

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

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

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

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

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.

Problem: Partial melting

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

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:

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:

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?

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.

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