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UB, Phy101: Chapter 13, Pg 1 Physics 101: Chapter 13 The Transfer of Heat l Textbook Chapter 13 è The Transfer of Heat »convection »conduction »Radiation.

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Presentation on theme: "UB, Phy101: Chapter 13, Pg 1 Physics 101: Chapter 13 The Transfer of Heat l Textbook Chapter 13 è The Transfer of Heat »convection »conduction »Radiation."— Presentation transcript:

1 UB, Phy101: Chapter 13, Pg 1 Physics 101: Chapter 13 The Transfer of Heat l Textbook Chapter 13 è The Transfer of Heat »convection »conduction »Radiation

2 UB, Phy101: Chapter 13, Pg 2

3 UB, Phy101: Chapter 13, Pg 3 Chapter 13, Preflight Which of the following is an example of convective heat transfer? 1. You stir some hot soup with a silver spoon and notice that the spoon warms up. 2. You stand watching a bonfire, but can’t get too close because of the heat. 3. Its hard for central air-conditioning in an old house to cool the attic.

4 UB, Phy101: Chapter 13, Pg 4 Chapter 13, Preflight Which of the following is an example of conductive heat transfer? 1. You stir some hot soup with a silver spoon and notice that the spoon warms up. 2. You stand watching a bonfire, but cant get too close because of the heat. 3. Its hard for central air-conditioning in an old house to cool the attic.

5 UB, Phy101: Chapter 13, Pg 5 Chapter 13, Preflight Which of the following is an example of radiative heat transfer? 1. You stir some hot soup with a silver spoon and notice that the spoon warms up. 2. You stand watching a bonfire, but cant get too close because of the heat. 3. Its hard for central air-conditioning in an old house to cool the attic.

6 UB, Phy101: Chapter 13, Pg 6 Heat Transfer: Convection l Air heats at bottom l Thermal expansion…density gets smaller l Lower density air rises è Archimedes: low density floats on high density l Cooler air pushed down l Cycle continues with net result of circulation of air l Practical aspects è heater ducts on floor è A/C ducts on ceiling è stove heats water from bottom heater

7 UB, Phy101: Chapter 13, Pg 7

8 UB, Phy101: Chapter 13, Pg 8 Heat Transfer: Conduction l Hot molecules have more KE than cold molecules l High-speed molecules on left collide with low-speed molecules on right è energy transferred to lower-speed molecules è heat transfers from hot to cold l H = rate of heat transfer = Q/t [J/s] è H = k A (T H -T C )/L »Q/t = k A  T/  x è k = “thermal conductivity” »Units: J/s-m-C »good thermal conductors…high k »good thermal insulators … low k T H Hot T C Cold L =  x Area A

9 UB, Phy101: Chapter 13, Pg 9 Chapter 13, Preflight On a cool night you make your bed with a thin cotton sheet covered by a thick wool blanket. As you lay there all covered up, heat is leaving your body, flowing though the sheet and the blanket and into the air of the room. (The trick, of course, is to have enough blankets to make the flow of heat just right...if not enough flows you will be too hot and if too much flows you will be cold). Compare the amount of heat that flows though the sheet to the amount of heat that flows through the blanket. 1. More heat flows through sheet than through the blanket. 2. More heat flows through blanket than through the sheet. 3. The same amount of heat flows through sheet and through the blanket. Sort of like the plywood and insulation thing in the book...it's not like the area between the sheet and blanket is getting hotter and hotter or colder and colder, so the same amount must flow through each. Also, kind of like the whole "hose doesn't leak" thing, I guess. correct

10 UB, Phy101: Chapter 13, Pg 10 l Example with 2 layers: find H=Q/t in J/s è Key Point: Continuity (just like fluid flow) »H 1 = H 2 »k 1 A(T 0 -T C )/  x 1 = k 2 A(T H -T 0 )/  x 2 » solve for T 0 = temp. at junction »then solve for H 1 or H 2 n answers: T 0 =2.27 C H=318 Watts  x 1 = 0.02 m A 1 = 35 m 2 k 1 = 0.080 J/s-m-C  x 2 = 0.075 m A 1 = 35 m 2 k 1 = 0.030 J/s-m-C Inside: T H = 25C Outside: T C = 0C H1H1 H2H2 T0T0

11 UB, Phy101: Chapter 13, Pg 11 Heat Transfer: Radiation l All things radiate electromagnetic energy è H emit = Q/t = eA  T 4 »e = emissivity (between 0 and 1) n perfect “black body” has e=1 »T is Kelvin temperature »  = Stefan-Boltzmann constant = 5.67 x 10 -8 J/s-m 2 -K 4 è No “medium” required l All things absorb energy from surroundings è H absorb = eA  T 0 4 »good emitters (e close to 1) are also good absorbers T Surroundings at T 0 Hot stove

12 UB, Phy101: Chapter 13, Pg 12 Heat Transfer: Radiation l All things radiate and absorb electromagnetic energy è H emit = Q/t = eA  T 4 è H absorb = eA  T 0 4 è H net = H emit - H absorb = eA  (T 4 - T 0 4 ) »if T>T 0, object cools down »if T { "@context": "http://schema.org", "@type": "ImageObject", "contentUrl": "http://images.slideplayer.com/13/4043697/slides/slide_12.jpg", "name": "UB, Phy101: Chapter 13, Pg 12 Heat Transfer: Radiation l All things radiate and absorb electromagnetic energy è H emit = Q/t = eA  T 4 è H absorb = eA  T 0 4 è H net = H emit - H absorb = eA  (T 4 - T 0 4 ) »if T>T 0, object cools down »if TT 0, object cools down »if T

13 UB, Phy101: Chapter 13, Pg 13 Chapter 13, Preflight One day during the winter, the sun has been shining all day. Toward sunset a light snow begins to fall. It collects without melting on a cement playground, but it melts immediately upon contact on a black asphalt road adjacent to the playground. How do you explain this. the asphalt is closer to being a perfect blackbody. It absorbes more heat from the sun as well as emits this heat. That is why the snow melts. Black (asphalt) absorbs electromagnetic waves (radiation) more readily than white (cement) does. Hence, the black has more radiation to emit because it has absorbed more. As a result, it releases more radiation into the snow, causing the snow to heat up, and melt.


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