Physics 101: Lecture 28 The Transfer of Heat

1. Physics 101: Lecture 28The Transfer of Heat • Today’s lecture will cover Textbook Chapter 13 • The Transfer of Heat • Convection • Conduction • Radiation

2. Concept Question Which of the following is an example of convective,conductive andradiative 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.

3. Heat Transfer: Convection • Convection: heat is transferred by the bulk movement of a gas or liquid. • Example: Fluid is sitting on a burner is heated from below. The fluid right above the flame is getting hot and thus expands: V increases => density decrease Thus, the hotter fluid experiences a net force upward (buoyant force) FB = rcold V g > rhot V g Archimedes Principle: low density floats on high density => warmer fluid moves upward and is replaced by cooler fluid -> fluid is warmed -> moves upward -> and so on • Cycle continues with net result of circulation of fluid that carries heat. • Practical aspects: • heater ducts on floor • A/C ducts on ceiling • stove heats water from bottom

4. Heat Transfer: Conduction • Conduction: Heat is transferred directly trough a material (bulk motion does not play a role). • Atoms/molecules of hotter materials have more KE than atoms/molecules of cooler materials. • Gas/fluids: high-speed atoms/molecules collide with low-speed atoms/molecules: • energy transferred to lower-speed atoms/molecules • heat transfers from hot to cold Metals: electrons can “freely” move and can transport energy • P = rate of heat transfer = Q/t [J/s] • Q = k A (TH-TC)/L • k = “thermal conductivity” • Units: J/(s m C) • good thermal conductors…high k (e.g. metals) • good thermal insulators … low k (e.g. wood) L = Dx TC Cold TH Hot Area A

5. Example with 2 layers: find P=Q/t in J/s • Key Point: Continuity (just like fluid flow) • P1 = P2 • k1A(T0-TC)/Dx1 = k2A(TH-T0)/Dx2 • solve for T0 = temp. at junction • then solve for P1 or P2 • answers: T0=5.8 C P=265 Watts P1 P2 Inside: TH = 25C Outside: TC = 4C T0 Dx1 = 0.019 m A1 = 35 m2 k1 = 0.080 J/s-m-C Dx2 = 0.076 m A2 = 35 m2 k2 = 0.030 J/s-m-C

6. Surroundings at T0 T Hot stove Heat Transfer: Radiation • Radiation: All things radiate electromagnetic energy. • Pemit = Q/t = eAT4 • e = emissivity (between 0 and 1) • perfect “black body” has e=1 • T is Kelvin temperature •  = Stefan-Boltzmann constant = 5.67 x 10-8 J/(s m2K4) • No “medium” required • All things absorb electromagnetic energy from surroundings. • Pabsorb = eAT04 • good emitters (e close to 1) are also good absorbers

7. Heat Transfer: Radiation Surroundings at T0 T Hot stove • All things radiate and absorb electromagnetic energy. • Pemit = Q/t = eAT4 • Pabsorb = eAT04 • Pnet = Pemit - Pabsorb = eA(T4 - T04) • if T>T0, object cools down if T<T0, object heats up

8. NASA’s Thermal Imaging System • http://mars.jpl.nasa.gov/ During day time the sun heats the surface of Earth or Mars. At night the materials on the surface emit this thermal energy again in form of radiation in the infrared. Since different materials emit differently, a whole spectrum of infrared radiation is measured which is used to identify the material.

9. Concept Question 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. Black (asphalt) absorbs electromagnetic waves (radiation) more readily than white (cement) does (emissivity is larger). 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.