PHY 2311

PHYSICS 231Lecture 30: Convection, Radiation &

Thermodynamics

Remco ZegersQuestion hours: Thursday 12:00-13:00 & 17:15-

18:15Helproom

PHY 2312

Convection

T high low

PHY 2313

Radiation

Nearly all objects emit energy through radiation:

P=AeT4 : Stefan’s law (J/s)=5.6696x10-8 W/m2K4

A: surface areae: object dependent constant emissivity (0-1)T: temperature (K)

P: energy radiated per second.

If an object is at Temperature T and its surroundingsare at T0, then the net energy gained/lost is:P=Ae(T4-To

4)

PHY 2314

emissivity

Ideal reflectore=0no energy is absorbed

Ideal absorber (black body)e=1all energy is absorbedalso ideal radiator!

PHY 2315

A BBQ

The coals in a BBQ cover an area of 0.25m2. If the emissivity of the burning coal is 0.95 and their temperature 5000C, how much energy is radiated everyminute?

P=AeT4 J/s =5.67x10-8*0.25*0.95*(773)4=4808 J/s

1 minute: 2.9x105 J (to cook 1 L of water 3.3x105 J)

PHY 2316

Thermodynamics (chapter 12)

Piston is moved downwardslowly so that the gas remainsin thermal equilibrium:The temperature is the thesame at all times in the gas,but can change as a whole.

vin

voutVout>Vin

Work is done on the gas

PHY 2317

Isobaric compression

Let’s assume that the pressure does notchange while lowering the piston (isobariccompression).

W=-Fy=-PAy (P=F/A)W=-PV=-P(Vf-Vi) (in Joule)

W: work done on the gas+ if V<0- if V>0

This corresponds to the area underthe curve in a P-V diagram

PHY 2318

Non-isobaric compression

In general, the pressure can change whenlowering the piston.

The work done on the gas when goingfrom an initial state (i) to a final state (f) is the area under the P-V diagram.

The work done by the gas is theopposite of the work done on the gas.

PHY 2319

Work done on gas: signs.

If the arrow goes from right to left, positive work isdone on the gas. If the arrow goes from left to right, negative work is done on the gas (the gas has done positive work on the piston) Not mentioned in the book!

PHY 23110

question

v

P

v

P

v

P

In which case is the work done on the gas largest?

A

B C

The area under the curves in cases B and C is largest (I.e. the absolute amount of work is largest). In caseC, the volume becomes larger and the pressure lower(the piston is moved up) so work is done by the gas (workdone on the gas is negative). In case B the work done onthe gas is positive, and thus largest.

PHY 23111

Isovolumetric process

v

P

Work done on/by gas: W=-PV=0

PHY 23112

First Law of Thermodynamics

By performing work on an objectthe internal energy can be changed

By transferring heat to an objectthe internal energy can be changed

The change in internal energy depends on the workdone on the object and the amount of heat transferredto the object.

Remember: the internal energy is the energy associatedwith translational, rotational, vibrational motion of atomsand potential energy

Think about deformation/pressure

Think about heat transfer

PHY 23113

First Law of thermodynamics

U=Uf-Ui=Q+W

U=change in internal energyQ=energy transfer through heat (+ if heat is

transferred to the system)W=energy transfer through work (+ if work is

done on the system)

This law is a general rule for conservation of energy

PHY 23114

First law: examples 1) isobaric process

A gas in a cylinder is kept at 1.0x105 Pa. The cylinder isbrought in contact with a cold reservoir and 500 J of heatis extracted. Meanwhile the piston has sunk and the volumechanged by 100cm3. What is the change in internal energy?

Q=-500 JW=-PV=-1.0x105 x -100x10-6m=10 JU=Q+W=-500+10=-490 J

In an isobaric process both Q and W are non-zero.

PHY 23115

First Law: examples: 2) Adiabatic process

A piston is pushed down rapidly. Because thetransfer of heat through the walls takes along time, no heat can escape. During the moving of the piston, the temperature has risen 1000C. If the container contained 10 mol of an ideal gas, how much work has been done during the compression?

Adiabatic: No heat transfer, Q=0U=Q+W=WU=3/2nRT=3/2x10x8.31x100=12465 J (ideal gas: only internal energy is kinetic energy U=3/2nRT)12465 J of work has been done on the gas.Why can we not use W=-PV???

PHY 23116

First Law: examples 3) general case

V(m3)

P(x105 Pa)

1 4

3

6 In ideal gas is compressed (seeP-V diagram).A) What is the change in internal energyb) What is the work done on the gas?C) How much heat has been transferred to the gas?

A) Use U=3/2nRT and PV=nRT so, U=3/2PV & U=3/2(PV) U=3/2(PfVf-PiVi)=3/2[(6E+05)x1 - (3E+05)x4)=-9E+5 J

i

f

B) Work: area under the P-V graph: (9+4.5)x105=13.5x105

(positive since work is done one the gas)C) U=Q+W so Q=U-W=(-9E+5)-(13.5E+5)=-22.5E+5 J Heat has been extracted from the gas.

PHY 23117

Types of processes

A: Isovolumetric V=0B: Adiabatic Q=0C: Isothermal T=0D: Isobaric P=0

PV/T=constant

PHY 23118

Metabolism

U=Q+W

Change in internal energy: Must be increased: Food!

Heat transfer: Negativebody temperature < room temperature

Work done (negative)

PHY 23119

Metabolic rate

U = Q + Wt t t

Metabolic rate: rate in which food and oxygen are transformed into internal energy (to balance losses dueto heat loss and work done).

|W/t| Body’s efficiency: |U/t|

PHY 23120

Body’s efficiency

U/t~oxygen use ratecan be measured

W/t can be measured