Purdue University, Physics 220 Lecture 25 Heat Transfer Lecture 25 Purdue University, Physics 220

Purdue University, Physics 220 Heat transfer Conduction Convection Electro-magnetic radiation Lecture 25 Purdue University, Physics 220

Heat Transfer: Conduction Hot molecules have more KE than cold molecules High-speed molecules on left collide with low-speed molecules on right energy transferred to lower-speed molecules heat transfers from hot to cold vibrations Lecture 25 Purdue University, Physics 220

Heat Transfer: Conduction I = rate of heat transfer = Q/t [J/s] I = k A (TH-TC)/d Q/t = k A T/x k = thermal conductivity Units: J/s*m*C good conductors…high k e.g., metal good insulators … low ke.g., plastic R = d/(Ak) = thermal resistance d = Dx TH Hot TC Cold Area A Lecture 25 Purdue University, Physics 220

Purdue University, Physics 220 Conduction Which of the following is an example of conductive heat transfer? A) You stir some hot soup with a silver spoon and notice that the spoon warms up. B) You stand watching a bonfire, but can’t get too close because of the heat. C) Its hard for central air-conditioning in an old house to cool the attic. Lecture 25 Purdue University, Physics 220

Conduction with 2 Layers Find I=Q/t in J/s Key Point: Continuity (just like fluid flow) I1 = I2 k1A(T0-TC)/Dx1 = k2A(TH-T0)/Dx2 solve for T0 = temp. at junction then solve for I1 or I2 TH-T0 = I R1 and T0-TC = I R2 T = (TH-T0) + (T0-TC) = I (R1 + R2) Inside: TH = 25C Outside: TC = 0C I1 I2 T0 Do transparency w/ problem. Dx1 = 0.02 m A1 = 35 m2 k1 = 0.080 J/s*m*C Dx2 = 0.075 m A2 = 35 m2 k2 = 0.030 J/s*m*C answer: T0=2.27 C I=318 Watts Lecture 25 Purdue University, Physics 220

Purdue University, Physics 220 iClicker Touch the metal base of a chair and the top of a wooden desk in an air-conditioned room, which feels colder? A) Base B) Same C) Desk Both must be the same temperature (room temperature), but metal feels colder because it conducts heat better/faster. Lecture 25 Purdue University, Physics 220

Heat Transfer: Convection Air heats at bottom Thermal expansion…density gets smaller Lower density air rises Archimedes: low density floats on high density Cooler air pushed down Cycle continues with net result of circulation of air I = Q/t = h A T h = coefficient of convection Practical aspects heater ducts on floor A/C ducts on ceiling stove heats water from bottom Lecture 25 Purdue University, Physics 220

Purdue University, Physics 220 Convection Lecture 25 Purdue University, Physics 220

Purdue University, Physics 220 Convection Which of the following is an example of convective heat transfer? A) You stir some hot soup with a silver spoon and notice that the spoon warms up. B) You stand watching a bonfire, but can’t get too close because of the heat. C) Its hard for central air-conditioning in an old house to cool the attic. Lecture 25 Purdue University, Physics 220

Purdue University, Physics 220 Convection Lecture 25 Purdue University, Physics 220

Heat Transfer: Radiation All things radiate / absorb electromagnetic energy Why? - Because vibration or/and acceleration of charged particles (electrons, protons, ions) results in electromagnetic radiation and vice versa. No “medium” required How amount of emitted / absorbed radiation depends on body temperature? Lecture 25 Purdue University, Physics 220

Heat Transfer: Radiation Iemit = Q/t = eAT4 Stefan-Boltzmann Law e = emissivity (between 0 and 1) perfect “black body” has e=1 (What does it mean “black body”?) T is the temperature of the object (in Kelvin)  = Stefan-Boltzmann constant = 5.67 x 10-8 J/s*m2*K4 Iabsorb = eAT04 good emitters (e close to 1) are also good absorbers Lecture 25 Purdue University, Physics 220

Purdue University, Physics 220 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. The black asphalt absorbs more heat from the sun. The black asphalt has an emissivity of 1 and absorbs energy from the surrounding (from the sun) compared to a smaller number for the cement. As a result, it is at a higher temperature than the cement and melts the snow. Lecture 25 Purdue University, Physics 220

Heat Transfer: Radiation All things radiate and absorb electromagnetic energy Iemit = eAT4 Iabsorb = eAT04 Inet = Iemit - Iabsorb = eA(T4 - T04) if T > T0, object cools down if T < T0, object heats up T Surroundings at T0 Hot stove Lecture 25 Purdue University, Physics 220

Purdue University, Physics 220 Exercise The Earth has a surface temperature around 270 K and an emissivity of 0.8, while space has a temperature of around 2 K. What is the net power radiated by the earth into free space? (Radii of the Earth and the Sun are RE = 6.38×106 m, RS = 7×108 m.) Inet = Iemit - Iabsorb = eA(T4 - T04) Lecture 25 Purdue University, Physics 220

Purdue University, Physics 220 Radiation Which of the following is an example of radiative heat transfer? A) You stir some hot soup with a silver spoon and notice that the spoon warms up. B) You stand watching a bonfire, but cant get too close because of the heat. C) Its hard for central air-conditioning in an old house to cool the attic. Lecture 25 Purdue University, Physics 220

Purdue University, Physics 220 iClicker If you place your hand underneath, but not touching, a kettle of hot water (about 80C), you mainly feel the presence of heat from: A) conduction B) convection C) radiation Lecture 25 Purdue University, Physics 220

Purdue University, Physics 220 Radiation Spectrum How frequency of radiation depends on T? Infrared: 100 m - 0.7 m Visible Light: 0.7m-0.4m Ultraviolet: < 0.4 m (stars which are hotter than sun) Wien’s Law: maxT = 2.898 × 10-3 mK Lecture 25 Purdue University, Physics 220

Purdue University, Physics 220 Night vision goggles Lecture 25 Purdue University, Physics 220

Purdue University, Physics 220 Temperature of the Sun The Sun appear yellowish, i.e., = 0.5 m. What is the temperature at the surface of the Sun? m T = 2.898 x 10-3 m K T = 2.898 x 10-3 / 0.5 x 10-6 = 6000 K Lecture 25 Purdue University, Physics 220

Purdue University, Physics 220 Greenhouse Effect Lecture 25 Purdue University, Physics 220