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Todays class Sections 15-1 to 15-4

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Todays class Sections 15-1 to 15-4 Powered By Docstoc
					SPH4UAP Day 20 Notes Today’s class: Sections 15-1 to 15-4 This is the heavy chapter of this unit. Note that you need to have a good understanding of chapters 13 and 14 before you can really understand the new material here.

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Thermodynamics is the study of transfers of energy by work and heat. Work is a mechanical transfer of energy (a force acting through a distance), while heat is a transfer of energy due to a temperature difference. The first thing you have to have an understanding of is the word “system”. We used this word before, when we talked about momentum. A system is simply the collection of objects. Anything not in the system is referred to as the “environment”. Systems may be closed (only energy, and not matter are able to be exchanged with the system); open (matter and energy are able to be exchanged); or isolated (nothing is exchanged with the system). The First Law of Thermodynamics is really a statement of the Law of Conservation of Energy. It says that the net change of internal energy for any system depends on two things: the amount of work done, and the amount of heat transferred. Note that this makes sense, as these are the only two ways in which energy can be transferred. There is, however, the possibility of confusing the signs (+ or -) inside the equation form of this law, which is: ΔU = Q - W (Q = heat added TO system; W = work done BY system) As far as systems go, we will be mostly interested systems which consist of ideal gases. Note that gases can be 1. Heated or Cooled (changed in temperature), 2. expanded or contracted (changed in volume), 3. increased or decreased in pressure. These changes can go about in several ways, referred to as “processes” Process Isothermal Adiabatic Isobaric Isochoric Condition Temperature remains constant (heat may be added or removed to ensure this) W = Q Heat is not exchanged between system and environment; ie Q = 0 and so ΔU = -W Pressure is kept constant also known as isovolumetric, volume is held constant; W = 0

Note that a work is done on a sample of gas when it is compressed, and gas does work when it is allowed to expand. If the pressure is constant, then the work done by a gas is W = Fd = (PA)d = PΔV The Second Law of Thermodynamics is a law describing that some things never happen, even though they would not violate any other laws of physics, if they were

SPH4UAP Day 20 Notes

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to occur. It is actually a complicated law that we will revisit later, but for now, one aspect of it is to say that “heat flows naturally from a hot object to a cold object; heat will not flow spontaneously from a cold object to a hot object”. Homework: - Read Sections 15-1 to 15-4 - Questions 1-7 - Problems # 1-15 odd - Read ahead

Answers to Questions
1. If water vapor condenses on the outside of a cold glass of water, the internal energy of the water vapor has decreased, by an amount equal to the heat of vaporization of the water vapor. Heat energy has left the water vapor, causing it to condense, and heat energy has entered the glass of water, and the air, causing them to get slightly warmer. No work is done, but heat is exchanged. During compression, work is done on the gas. Assuming that there is no heat flow to or from the gas (since the process is quick), by conservation of energy (the first law of thermodynamics) the work done on the gas becomes internal energy of the gas, and so the temperature of the gas is increased. During expansion, work is done by the gas on its surroundings. Again assuming that there is no heat flow to or from the gas, by conservation of energy, the work is done by the gas at the expense of the internal energy of the gas, and so the temperature of the gas is decreased. Since the process is isothermal, there is no change in the internal energy of the gas. Thus U  Q  W  0  Q  W , and so the heat absorbed by the gas is equal to the work done by the gas. Thus 3700 J of heat was added to the gas. It is possible for temperature (and thus internal energy) to remain constant in a system even though there is heat flow into or out of the system. By the first law of thermodynamics, there must be an equal amount of work done on or by the system, so that U  Q  W  0  Q  W . The isothermal expansion or compression of a gas would be an example of this situation. If the gas is compressed adiabatically, no heat enters or leaves from the gas. The compression means that work was done ON the gas. By the first law of thermodynamics, U  Q  W , since Q  0 , then U  W . The change in internal energy is equal to the opposite of the work done by the gas, or is equal to the work done on the gas. Since positive work was done on the gas, the internal energy of the gas increased, and that corresponds to an increase in temperature. This is conservation of energy – the work done on the gas becomes internal energy of the gas particles, and the temperature increases accordingly. Mechanical energy can be transformed completely into heat. As a moving object slides across a rough level floor and eventually stops, the mechanical energy of the moving object has been transformed completely into heat. Also, if a moving object were to be used to compress a frictionless piston containing an insulated gas, the kinetic energy of the object would become internal energy of the gas. A gas that expands adiabatically (without heat transfer) transforms internal energy into mechanical energy, by doing work on its

2.

3.

4.

5.

6.

SPH4UAP Day 20 Notes
surroundings at the expense of its internal energy. Of course, that is an ideal (reversible) process. 7.

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It is possible to warm the kitchen in the winter by having the oven door open. The oven heating elements radiate heat energy into the oven cavity, and if the oven door is open, the oven is just heating a bigger volume than usual. However, you cannot cool the kitchen by having the refrigerator door open. The refrigerator exhausts more heat than it removes from the refrigerated volume, so the room actually gets warmer with the refrigerator door open. If you could have the refrigerator exhaust into some other room, then the refrigerator would be similar to an air conditioner, and it could cool the kitchen, while heating up some other space.

Solutions to Problems
1. The kcal is the heat needed to raise 1 kg of water by 1 Co. Use the definition to find the heat needed. 1 kcal  4186 J  7  30.0 kg  95o C  15o C   1.0 10 J o  1kg  1C  1 kcal 









3.

(a) (b)

 4.186 103 J  7   1.0 10 J  1 Cal   1 kWh  2500 Cal    2.9 kWh  860 Cal 
2500 Cal 

(c) At 10 cents per day, the food energy costs $0.29 per day . It would be practically impossible to feed yourself in the United States on this amount of money.

5.

The energy input is causing a certain rise in temperature, expressible as a number of Joules per hour per Co. Convert that to mass using the definition of kcal. o  3.2 107 J h   1 kcal  1kg  1C 2  35 Co   4186 J  1 kcal  2.2 10 kg h   

 

7.

The energy generated by using the brakes must equal the car’s initial kinetic energy, since its final kinetic energy is 0.

  1m s    1 kcal  2 Q  mv  1.2  10 kg   95 km h      4186 J   1.0  10 kcal   3.6 km h    
1 2 2 0 1 2 3

2

9.

The specific heat can be calculated from Eq. 14-2.

Q  mcT  c 

Q mT



 5.1 kg   31.5 C  18.0
o

1.35  105 J

o

C



 1961 J kg Co  2.0  103 J kg Co

SPH4UAP Day 20 Notes

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11. The heat gained by the glass thermometer must be equal to the heat lost by the water.
mglass cglass Teq  Ti glass   mH O cH O Ti H O  Teq
2 2 2





 35 g   0.20 cal
Ti H O  40.1o C
2

g Co

39.2 C  21.6 C   135 g  1.00 cal g C  T
o o o

i H2O

 39.2o C



13. The heat lost by the horseshoe must be equal to the heat gained by the iron pot and the water. Note that 1 L of water has a mass of 1 kg.

mshoe cFe Ti shoe  Teq  mpot cFe Teq  Ti pot  mH O cH O Teq  Ti H O

 0.40 kg   450 J





kg C

o

 T



i shoe

    25.0 C    0.30 kg   450 J kg C  25.0C  20.0C   1.35 kg   4186 J kg C  25.0C  20.0C 
2 2 2

o

o

o

o

o

o

o

Ti shoe  186o C  190o C

15. The heat must warm both the water and the pot to 100oC. The heat is also the power times the time.

Q  Pt  mAl cAl  mH O cH O TH O 
2 2 2





m t

Al Al

c  mH O cH O TH O
2 2 2



P

 0.36 kg   900 J kg Co    0.75 kg   4186 J kg Co    92Co   
750 W

 425 s  7 min


				
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