1. PHY 231 1 PHYSICS 231 Lecture 26: Conduction,Convection & Radiation Remco Zegers walk-in: Tue 4-5 Helproom

2. PHY 231 2 Heat transfer to an object Q=cmTQ=cmT Energy transfer (J or cal) Specific heat (J/(kg o C) or cal/(g o C) Mass of object Change in temperature The amount of energy transfer Q to an object with mass m when its temperature is raised by  T:

3. PHY 231 3 Phase Change GAS(high T) liquid (medium T) Solid (low T) Q=c gas m  T Q=c liquid m  T Q=c solid m  T Gas  liquid liquid  solid Q=mL f Q=mL v

4. PHY 231 4 question Given: L f =6.44x10 4 J/kg T melt =1063 o C c specific =129 J/kg 0 C A block of gold (room temperature 20 0 C) is found to just melt completely after supplying 4x10 3 J of heat. What was the mass of the gold block? a)0.01 kg b)0.02 kg c)0.03 kg d)0.06 kg e)10 kg Q=cm  T+mL f =129*m*1043+m*6.44x10 4 =2x10 5 *m 4000=2x10 5 m m=0.02 kg

5. PHY 231 5 phase transformations depend on pressure as well. Not treated in the book (not part of material for exam).

6. PHY 231 6 quiz Ice is heated steadily and becomes liquid and then vapor. During this process: a)the temperature rises continuously. b)when the ice turns into water, the temperature drops for a brief moment. c) the temperature is constant during the phase transformations d) the temperature cannot exceed 100 o C

7. PHY 231 7 How can heat be transferred?

8. PHY 231 8 Conduction Touching different materials: Some feel cold, others feel warm, but all are at the same temperature…

9. PHY 231 9 Thermal conductivity metal wood T=37 0 C T=20 0 C The heat transfer in the metal is much faster than in the wood: (thermal conductivity)

10. PHY 231 10 Heat transfer via conduction Conduction occurs if there is a temperature difference between two parts of a conducting medium Rate of energy transfer P P=Q/  t (unit Watt) P=kA(T h -T c )/  x=kA  T/  x k: thermal conductivity Unit:J/(ms o C) metal k~300 J/(ms o C) gases k~0.1 J/(ms o C) nonmetals~1 J/(ms o C)

11. PHY 231 11 Example A glass window (A=4m 2,  x=0.5cm) separates a living room (T=20 0 C) from the outside (T=0 o C). A) What is the rate of heat transfer through the window (k glass =0.84 J/(ms o C))? B) By what fraction does it change if the surface becomes 2x smaller and the temperature drops to -20 0 C? A) P=kA  T/  x=0.84*4*20/0.005=13440 Watt B) P orig =kA  T/  x P new =k(0.5A)(2  T)/  x=P orig The heat transfer is the same

12. PHY 231 12 Another one. Heat reservoir Heat sink An insulated gold wire (I.e. no heat lost to the air) is at one end connected to a heat reservoir (T=100 0 C) and at the other end connected to a heat sink (T=20 0 C). If its length is 1m and P=200W what is its cross section (A)? k gold =314 J/(ms 0 C). P=kA  T/  x=314*A*80/1=25120*A=200 A=8.0E-03 m 2

13. PHY 231 13 And another Water 0.5L 100 0 C 150 0 C A=0.03m 2 thickness: 0.5cm. A student working for his exam feels hungry and starts boiling water (0.5L) for some noodles. He leaves the kitchen when the water just boils.The stove’s temperature is 150 0 C. The pan’s bottom has dimensions given above. Working hard on the exam, he only comes back after half an hour. Is there still water in the pan? (L v =540 cal/g, k pan =1 cal/(ms 0 C) To boil away 0.5L (=500g) of water: Q=L v *500=270000 cal Heat added by the stove: P=kA  T/  x=1*0.03*50/0.005= =300 cal P=Q/  t  t=Q/P=270000/300=900 s (15 minutes) He’ll be hungry for a bit longer…

14. PHY 231 14 Isolation L1L1 L2L2 L3L3 inside ThTh TcTc A house is built with 10cm thick wooden walls and roofs. The owner decides to install insulation. After installation the walls and roof are 4cm wood+2cm isolation+4cm wood. If k wood =0.10 J/(ms 0 C) and k isolation =0.02 J/(ms 0 C), by what factor does he reduce his heating bill? P before =A  T/[0.10/0.10]=A  T P after =A  T/[0.04/0.10+0.02/0.02+0.04/0.10]=0.55A  T Almost a factor of 2 (1.81)!

15. PHY 231 15 Convection T high  low

16. PHY 231 16 Radiation Nearly all objects emit energy through radiation: P=  AeT 4 : Stefan’s law (J/s)  =5.6696x10 -8 W/m 2 K 4 A: surface area e: object dependent constant emissivity (0-1) T: temperature (K) P: energy radiated per second.

17. PHY 231 17 emissivity Ideal reflector e=0 no energy is absorbed Ideal absorber (black body) e=1 all energy is absorbed also ideal radiator!

18. PHY 231 18 A BBQ The coals in a BBQ cover an area of 0.25m 2. If the emissivity of the burning coal is 0.95 and their temperature 500 0 C, how much energy is radiated every minute? P=  AeT 4 J/s =5.67x10 -8 *0.25*0.95*(773) 4 =4808 J/s 1 minute: 2.9x10 5 J (to cook 1 L of water 3.3x10 5 J)

19. PHY 231 19 radiation balance If an object would only emit radiation it would eventually have 0 K temperature. In reality, an object emits AND receives radiation. P=  Ae(T 4 -T 0 4 ) where T: temperature of object T 0 : temperature of surroundings.

20. PHY 231 20 example The temperature of the human body is 37 0 C. If the room temperature is 20 0 C, how much heat is given off by the human body to the room in one minute? Assume that the emissivity of the human body is 0.9 and the surface area is 2 m 2. P=  Ae(T 4 -T 0 4 ) =5.67x10 -8 * 2 * 0.9 *(310.5 4 -293.5 4 )= =185 J/s Q=P*  T=185*60=1.1x10 4 J