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lec20 - − T C 8.044 L20B3 55 o F subsurface temp at 40 o...

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MIT OpenCourseWare http://ocw.mit.edu 8.044 Statistical Physics I Spring 2008 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms .
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Refrigerator Run cycle backwards, extract heat at cold end, dump it at hot end HEAT EXTRACTED (COLD END) = | Q C | = | Q C | WORK DONE ON SUBSTANCE W | Q H | − | Q C | For the special case of a quasi-static Carnot cycle T C = T H T C 8.044 L20B1
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As with engine, can show Carnot cycle is optimum. Practical: increasingly difficult to approach T = 0. Philosophical: T = 0 is point at which no more heat can be extracted. 8.044 L20B2
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Heat Pump Run cycle backwards, but use the heat dumped at hot end. HEAT DUMPED (HOT END) = | Q H | = | Q H | WORK DONE ON SUBSTANCE W | Q H | − | Q C | For the special case of a quasi-static Carnot cycle
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Unformatted text preview: − T C 8.044 L20B3 55 o F subsurface temp. at 40 o latitude T C = 286 K → 70 o F room temperature T H = 294 K → 294 | Q H | ∼ 37 ∆ W ≤ 8 8.044 L20B4 3 rd law lim S = S 0 T 0 → At T = 0 the entropy of a substance approaches a constant value, independent of the other thermody-namic variables. • Originally a hypothesis • Now seen as a result of quantum mechanics Ground state degeneracy g (usually 1) S k ln g (usually 0) ⇒ → 8.044 L20B5 ± ² ∂S Consequences = 0 ∂x T =0 Example: A hydrostatic system 1 ± ∂V ² 1 ± ∂S ² α ≡ V ∂T P = − V ∂P T → 0 a s T → 0 V T α 2 C P − C V = K T → 0 a s T → 0 S ( T ) − S (0) = ³ T T =0 C V ( T T ± ± ) dT ± ⇒ C V ( T ) → 0 a s T → 0 8.044 L20B6...
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