1. Conduction C ontents: Basic Concept Example Whiteboards

2. Conduction Heat transfer by contact Random wiggling Hot to cold Flow rate   T, A, material, 1/thickness Simulation Insulation prevents conduction Down/Synthetics Thickness Demo – Ice Melter blocks. Heat sinks

3. Conduction ΔQ / t = Heat flow rate in J/s or Watts k = Thermal Conductivity (Wm -1o C -1 ) A = Area (m 2 ) l = Thickness of insulation (m) T 1 = High temperature ( o C) T 2 = Low Temperature ( o C)

6. A = 1.2x1.2 + 4x1.2x.95 = 6.00 m2 100 = k(6.00 m2)(43 oC)/0.12 k = 0.0465 W/m/oC 0.0465 W/m/ o C Example #1 – A 100. W light bulb can maintain a temperature of 15.0 o C in a well head enclosure when it is -28.0 o C outside. The enclosure is a box is 1.20 m wide x 1.20 m long x 0.950 tall insulated by 12.0 cm of insulation. What is the thermal conductivity of the insulation? (ignore the heat that goes into the ground)

7. Conduction 1-3

8. Q/t = (0.84)(1.2x2.4)(12-3.4)/(0.00400) = 5201.28 W 5200 W What is the heat flow rate through a 2.4 m x 1.2 m window that is a single pane when the air next to the glass is at 12.0 o C and the air outside is at 3.4 o C? The glass is 4.0 mm thick, and has a thermal conductivity of 0.84 W/m/ o C.

9. 140 = (0.048)(2.4x3.7)(ΔT)/(0.0894) ΔT = 29.3637387387 o C 29 o C What is the temperature difference between the inside and the outside of a wall if it is insulated by 8.94 cm of fiberglass insulation (k = 0.048 W/m/ o C) is 2.4 m tall by 3.7 m wide and heat is flowing through it at a rate of 140 J/s?

10. (0.175)*3.33E5/t = (0.84)(0.025)(20)/(0.0014) t = 194.25 s 190 s A glass (k = 0.84 J/m/ o C) beaker is 1.4 mm thick, contains 175 grams of ice (L = 3.33E5 J/kg) in an ice bath (at 0 o C of course), has a surface area of 0.025 m 2 exposed to a well stirred water bath maintained at 20.0 o C. What time will it take the ice to all melt? (Assume it has a perfectly insulating lid)