1 physics 1 thermal physics links: and //homepage.mac.com/phyzman/phyz/bop/2-06adht/index.html

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Physics 1

Thermal Physics

Links: http://homepage.mac.com/phyzman/phyz/BOP/2-06ADHT/index.html and http://www.colorado.edu/physics/2000/index.pl

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Assignment

• P&P12: • In class:1(a),3,4(a,b),• 6,19,22,23,32,58,59• 72oF tooC and -10oC to oF• HW:1(b-d),5,6,8,20,24,37,40,41,61,63,66•  Convert 25oF to oC and -210oC to oF• Link:

http://homepage.mac.com/phyzman/phyz/BOP/2-06ADHT/index.html

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What do you think? know?

1. Why does popcorn pop?

2. On a camping trip, your friend tells you that fluffing up a down sleeping bag before you go to bed will keep you warmer than sleeping in the same bag when it is still crushed. Why?

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3. Why is it difficult to build a fire with damp wood?

4. Why does steam at 100oC cause more severe burns than liquid water does at 100oC?

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5. Until refrigerators were invented, many people stored fruits and vegetables in underground cellars. Why was this more effective than keeping them in the open air?

6. In the past, when a baby had a high fever, the doctor might have suggested gently sponging off the baby with rubbing alcohol. Why would this help?

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7. Why does water expand when it freezes?

8. Why, during the final construction of the St. Louis arch, was water sprayed on the previous sections as the last section was put in place?

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Objectives

• 1. Describe thermal energy and compare it to potential and kinetic energies.

• 2. Describe changes in temperatures of two objects reaching thermal equilibrium

• 3. Identify various temperature scales, and convert between them

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Objectives

• 4. Explain heat as energy transferred between substances at different temperatures in one of 3 ways (conduction, convection, radiation)

• 5. Relate heat and temperature

• 6. Apply principle of energy conservation to calculate changes in potential, kinetic, & internal energy

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Objectives

• 7. Perform calculations with specific heat capacity

• 8. Interpret the various sections of a heating curve

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Objectives

• 9. Recognize that a system can absorb or release energy as heat in order for work to be done on or by that system

• 10. Compute work done during thermodynamic process

• 11. Distinguish between isovolumetric, isothermal, and adiabatic thermodynamic processes

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Objectives

• 12. Illustrate how the first law of thermodynamics is a statement of energy conservation

• 13. Calculate heat, work, and change in internal energy using lst law of T-D

• 14. Apply 1st law of T-D to describe cyclic processes

• 15. Recognize why 2nd law of T-D requires 2 bodies at different temps.

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1.Relate temperature to the kinetic energy of atoms and molecules

• Temperature scales• In the USA, the Fahrenheit temperature scale is used.

Most of the rest of the world uses Celsius, and in science it is often most convenient to use the Kelvin scale.

• The Celsius scale is based on the temperatures at which water freezes and boils. 0°C is the freezing point of water, and 100° C is the boiling point. Room temperature is about 20° C, a hot summer day might be 40° C, and a cold winter day would be around -20° C.

• To convert between Fahrenheit and Celsius, use these equations:

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• Convert 72 oF to oC

• C = 5/9 (F-32)

• C = 5/9 (72-32) = 22oC

• Convert -10 oC to oF

• F = 9/5 C + 32

• F = 9/5(-10) + 32 = 14oF

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Temperature degree scales comparison 

Objective 3:

Temperature Scales

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Celsius to Kelvin:

T = Tc + 273.15Problem:

1. The lowest outdoor temperature ever recorded on Earth is -128.6 oF., recorded at Vostok Station, Antarctica, in 1983. What is this temperature on the Celsius and Kelvin scales?Answers: -89.22oC, 183.93 K

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• Obj. #2 - Hotter temperature means

• – more heat present in a substance

• – the faster the molecules of the substance move.

• Obj #5 – Relate heat and temperature

• Heat units: calorie or joule (amount of heat energy present in a substance)

• Temperature units: degree (proportional to heat energy present in a substance)

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2.Describe changes in temperatures of two objects reaching thermal equilibrium

• The temperature of the hotter substance will decrease. The temperature of the colder substance will increase. Each change will stop when the temperatures are the same – thermal equilibrium. In other words, thermal energy travels from hot to cold.

• Obj. #4 - Heat energy can be transferred by– Convection (motion of fluid),

conduction (touching), or radiation (electromagnetic waves)

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ConvectionHeat transfer in fluids generally takes place via

convection. Convection currents are set up in the fluid because the hotter part of the fluid is not as dense as

the cooler part, so there is an upward buoyant force on the hotter fluid, making it rise while the cooler, denser,

fluid sinks. Birds and gliders make use of upward convection currents to rise, and we also rely on

convection to remove ground-level pollution.

ConductionWhen heat is transferred via conduction, the substance itself does not flow; rather, heat is transferred internally, by vibrations of atoms and molecules. Electrons can also carry heat, which is the reason metals are generally very good conductors of heat.

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RadiationThe third way to transfer heat, in addition to convection and conduction, is by radiation, in which energy is transferred in the form of electromagnetic waves. More about electromagnetic waves in a lot more detail in a later chapter; an e-m wave is basically an oscillating electric and magnetic field traveling through space at the speed of light.

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Specific heat

The amount of energy that must be added to raise the temperature of a unit mass of a substance by one temperature unit.

• Units: J/kg K

• For Water: 4180 J/kg K

• For Aluminum: 897 J/kg K

• Which one heats faster?

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A similar table is on page 413 in the Hewitt book. The units are J/g oC in that table.

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Heat Transfer Q = mCΔT = mC (Tf – Ti)

• Heat Transfer

• Q = mCΔT = mC (Tf – Ti)

• Q, quantity of heat in joule

• m, mass of substance in kg

• c, specific heat for water in 4186 j/kg K

• t, temperature in Celsius

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See table 12-1 on page 318

Find the amount of heat needed to change the temperature of 5.0 g of liquid water from 8.0oC to 100oC.

Q = mct = .005kg(4186 j/kgoC) (92oC) = 1.9 x 103 j

Again,

specific heat is the amount of heat necessary to change one kg of a substance 1 degree Celsius or Kelvin.

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• 12/3 When you turn on the hot water to wash dishes, the water pipes have to heat up. How much heat is absorbed by a copper water pipe with a mass of 2.3 kg when its temperature is raised from 20.0oC to 80.0oC?

• Q = mct

• Q = (2.3kg)(390J/kgoC)(60.0oC)

• Q = 53820 J or 5.4x104J

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Temperature vs Heat

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12/4b

Melting or Freezing

• There is no temperature change. To determine heat lost of gained use:

• Q = mHf Hf is heat of fusion

• 3.33x105 J/kg for water

• How much heat must be absorbed to melt 5.0 g of ice at 0.0 oC?

• Q=mHf = 0.005kg x 3.33 105Jkg-1

• Q = 1.7 x 102 J of heat absorbed30

Evaporation or Condensation

• Again, no temp. change. Use Q = mHv

• For water Hv = 2.26 x 106 J/kg.

• How much heat is lost when 100.g of steam condenses at 100oC?

• Q = mHv = 0.100kg (2.26 x 106 J/kg)

• Q = 2.26 x 105 J heat lost

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> One step…

• How much heat is needed to raise the temperature of 50.0 g of ice at 0oC and change it to steam at 100.oC?

• Melt + raise temp + evap• Q = mHf + mct + mHv

• Q = 0.050kg[3.33 105Jkg-1 + (4186 j/kg-1oC-1)(100oC)+

• +(2.26 x 106 J/kg)]• Q = 3.01 x 106 J of heat absorbed.

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Your assignment:

• Find the amount of heat involved:

• #1 to heat 75.3g of water from 20.0oC and change it to steam at 100.oC.

• #2 to cool 124g of water from 100.oC and change it to ice at OoC.

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Specific Heat Capacities

Which

one is

greatest

that you

use

everyday?

OR J/kgoC

Law of Heat Exchange

• Heat lost = heat gained within a closed system.

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Law of Heat Exchange Q lost = Q gained

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• 0.300 kg of coffee, at a temperature of 95 °C, is poured into a room-temperature (20.oC) steel mug, of mass 0.125 kg. Assuming no energy is lost to the surroundings, what does the temperature of the mug filled with coffee come to?

• Applying conservation of energy, the total change in energy of the system must be zero. So, we can just add up the individual energy changes (the Q's) and set the sum equal to zero. The subscript c refers to the coffee, and m to the mug.

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Note that room temperature in Celsius is about 20°. Re-arranging the equation to solve for the final temperature gives:

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The temperature of the coffee doesn't drop by much because the specific heat of water (or coffee) is so much larger than that of steel. This is too hot to drink, but if you wait, heat will be transferred to the surroundings and the coffee will cool.

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Latent Heat is energy transferred during phase changes

• Latent Heat• Crystalline materials change phase -- melt and freeze or vaporize

and condense -- at a single, fixed temperature.Energy is required for a change of phase of a substance. The ratio of the energy to the mass of the substance involved is called the latent heat of the substance. This is much like the specific heat we have just discussed. It is called latent heat because there is no change or difference in temperature.

•Latent heat of fusion Hf describes the heat necessary to melt (or freeze) a unit mass of a substance.

Q = m Hf• Latent heat of vaporization Hv describes the heat necessary to

vaporize (or condense) a unit mass of a substance.

• Q = m Hv•

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Temperature vs Heat

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Formulas

• Temperature change use: Q = mc t• Melting or freezing: Q = m Hf

• Evaporation or Condensation: Q = mHv

• Hf is latent heat of fusion, 3.33 x 105 J/kg• Hv is latent heat of vaporization, 2.26 x 106

J/kg• These values are for water.

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Problem

• A jar of tea is placed in sunlight until it reaches an equilibrium temperature of 32oC. In an attempt to cool the liquid to 0oC, which has a mass of l80 g, how much ice at 0oC is needed? Assume the specific heat capacity of the tea to be that of pure liquid water.

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m tea = 180g m ice = ? which is the mass of the water that has melted

c tea = c water = 4186 J/kgoC

H f = 3.33 x 105 J/kg and t final = 32oC

Q lost = Q gained tea loses and water gains only melting the ice

(mct)tea = (mHf)ice

m ice = (mct)tea Hf ice

m ice = (.180kg)(4186 J/kgoC)(32oC)

3.33x105 J/kg

= 7.2 x 10-2kg

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1st Law of Thermodynamics

U = Q - W

The change in thermal energy of an object is equal to the heat added to the object minus the work done by the object.

See Fig 12-11 on

Page 326

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A heat engine

Transforms heat at high temperature into mechanical energy and low-temperature waste heat

A heat pump (refrigerator) absorbs heat from the cold reservoir and gives off heat to the hot reservoir.

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2nd Law of Thermodynamics

• In the 19th Century French engineer Sadi Carnot studied the ability of engines to convert thermal energy into mechanical energy.

• He developed a logical proof that even an ideal engine would generate some waste heat.

• Carnot’s result is best described by the term entropy which is the measure of the disorder in a system.

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The change is entropy, D S, is shown by the equation:

S = Q / T Units: J/K

The change in entropy of an object is equal to the heat added to the object divided by the temperature of the object.

Natural processes occur in a direction that increases the entropy of the universe. All processes tend toward disorder unless some action occurs to keep them ordered. i.e., heat can flow only from hot to cold naturally.

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Practice Problems

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12/32 answer

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•The End

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