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PHYSICS 231 Lecture 28: Conduction,Convection & Radiation

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1 PHYSICS 231 Lecture 28: Conduction,Convection & Radiation
Remco Zegers walk-in: Monday 9:15-10:15 Helproom PHY 231

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

3 Phase Change GAS(high T) Q=cgasmT Q=csolidmT Solid (low T) liquid 
Q=mLv liquid (medium T) Q=mLf Q=cliquidmT PHY 231

4 quiz (extra credit) A block of gold (room temperature 200C)
is found to just melt completely after supplying 4x103 J of heat. What was the mass of the gold block? Given: Lf=6.44x104 J/kg Tmelt=1063oC cspecific=129 J/kg0C 0.01 kg 0.02 kg 0.03 kg 0.06 kg 10 kg Q=cmT+mLf =129*m*1043+m*6.44x104 =2x105*m 4000=2x105m m=0.02 kg PHY 231

5 How can heat be transferred?
PHY 231

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

7 Thermal conductivity metal T=200C wood T=200C The heat transfer
in the metal is much faster than in the wood: (thermal conductivity) T=370C T=370C PHY 231

8 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(Th-Tc)/x=kAT/x k: thermal conductivity Unit:J/(msoC) metal k~300 J/(msoC) gases k~0.1 J/(msoC) nonmetals~1 J/(msoC) PHY 231

9 Example A glass window (A=4m2,x=0.5cm)
separates a living room (T=200C) from the outside (T=0oC). A) What is the rate of heat transfer through the window (kglass=0.84 J/(msoC))? B) By what fraction does it change if the surface becomes 2x smaller and the temperature drops to -200C? A) P=kAT/x=0.84*4*20/0.005=13440 Watt B) Porig=kAT/x Pnew=k(0.5A)(2T)/x=Porig The heat transfer is the same PHY 231

10 Another one. Heat sink Heat reservoir
An insulated gold wire (I.e. no heat lost to the air) is at one end connected to a heat reservoir (T=1000C) and at the other end connected to a heat sink (T=200C). If its length is 1m and P=200W what is its cross section (A)? kgold=314 J/(ms0C). P=kAT/x=314*A*80/1=25120*A=200 A=8.0E-03 m2 PHY 231

11 And another Water 0.5L 1000C A=0.03m2 thickness: 0.5cm. 1500C
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 1500C. 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? (Lv=540 cal/g, kpan=1 cal/(ms0C) To boil away 0.5L (=500g) of water: Q=Lv*500= 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… PHY 231

12 Isolation Tc Th inside L1 L2 L3
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 kwood=0.10 J/(ms0C) and kisolation=0.02 J/(ms0C), by what factor does he reduce his heating bill? Pbefore=AT/[0.10/0.10]=AT Pafter=AT/[0.04/ / /0.10]=0.55AT Almost a factor of 2 (1.81)! PHY 231

13 Convection T high  low PHY 231

14 Radiation Nearly all objects emit energy through radiation:
P=AeT4 : Stefan’s law (J/s) =5.6696x10-8 W/m2K4 A: surface area e: object dependent constant emissivity (0-1) T: temperature (K) P: energy radiated per second. PHY 231

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

16 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 every minute? 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 231

17 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(T4-T04) where T: temperature of object T0: temperature of surroundings. PHY 231

18 example The temperature of the human body is 370C. If the
room temperature is 200C, 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 m2. P=Ae(T4-T04) =5.67x10-8 * 2 * 0.9 *( )= =185 J/s Q=P*T=185*60=1.1x104 J PHY 231


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