Fa08-MT1-Smoot-Soln-1

Fa08-MT1-Smoot-Soln-1 - Physics 7B Fall 2008: Lecture 3...

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Unformatted text preview: Physics 7B Fall 2008: Lecture 3 Midterm 1, Prob 4 Data A = 25 cm 2 T h = 100 ◦ C T c = 0 ◦ C L Cu = 26 . cm L Al = 33 . cm k Cu = 400 W m · ◦ C k Al = 240 W m · ◦ C (a) Interface temperature The heat flows through each section must be equal to each other and to the overall heat flow through the entire composite bar at steady state. If they aren’t equal, there would be a net heat dump or removal at the interface, which would raise or lower the temperature there (which wouldn’t be steady state). Let the temperature at the interface be T. Then, the equations for heat conduction become: dQ dt k Cu = dQ dt k Al at steady-state. k Cu A ( T h- T ) L Cu = k Al A ( T- T c ) L Al (1) k Cu L Cu T h + k Al L Al T c = T k Cu L Cu + k Al L Al Finally, T = k Cu L Cu T h + k Al L Al T c k Cu L Cu + k Al L Al (2) Numerically, T = 400 W/m · ◦ C . 26 m 100 ◦ C + 240 W/m · ◦ C . 33 m ◦ C 400 W/m · ◦ C . 26 m + 240 W/m · ◦ C . 33 m T = 68 ◦ C = 3 . 4 × 10 2 K (3) (2 sig figs) Grading scheme: 1pt: Showing that you know WHY the two heat flows are equal (steady state), SOMEWHERE in the writeup. 1pt: Correct order of the 2 temperature differences 1pt: final symbolic answer 2pt: final numeric answer. Work not necessary (but useful for partial credit here). 1 (b) For T interface = 50 ◦ C We can directly rearrange eq. 1 to get L Al : L Al = k Al k Cu ( T- T c ) ( T h- T ) L Cu (4) Numerically, for the interface temperature T = 50 ◦ C : L Al = 240 W/m · ◦ C 400 W/m · ◦ C (50 ◦ C- ◦ C ) (100 ◦ C- 50 ◦ C ) 26 cm = 15 . 6 cm ≈ 16 cm (5) (2 sig figs) Grading scheme: 1pt: Correct order of the 2 temperature differences 2pt: final symbolic answer. Partial credit given depending on how much numerical work was shown. 2pt: final numeric answer. Work not necessary (but useful for partial credit here). (c) Heat flow along the original rods Again, we can use the original eq. 1 directly (any one of the 2 expressions is fine): dQ dt k Al = dQ dt k Cu = k Cu A ( T h- T ) L Cu (6) Numerically, dQ dt k Al = dQ dt k Cu = (400 W/m · ◦ C )(25 × 10- 4 m 2 )(100 ◦ C- 67 . 9 ◦ C ) . 26 m = 123 . 5 W ≈ . 12 kW (7) (2 sig figs) Grading scheme: 3pt: final symbolic answer (any one expression, with proper quantities - NOT generic T’s, L’s, etc.) Again, a LOT of explicit numeric plugging-in gave you some partial credit here....
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This note was uploaded on 02/04/2010 for the course PHYSICS 7B taught by Professor Packard during the Spring '08 term at Berkeley.

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Fa08-MT1-Smoot-Soln-1 - Physics 7B Fall 2008: Lecture 3...

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