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

Info iconThis preview shows pages 1–7. Sign up to view the full content.

View Full Document Right Arrow Icon
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 .
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
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
Background image of page 2
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
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
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 T H = T H
Background image of page 4
Background image of page 5

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 6
Background image of page 7
This is the end of the preview. Sign up to access the rest of the document.

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 S Consequences = 0 x T =0 Example: A hydrostatic system 1 V 1 S V T P = V P T 0 a s T 0 V T 2 C P C V = K T 0 a s T 0 S ( T ) S (0) = T T =0 C V ( T T ) dT C V ( T ) 0 a s T 0 8.044 L20B6...
View Full Document

This note was uploaded on 11/08/2011 for the course PHYSICS 8.004 taught by Professor Staff during the Spring '08 term at MIT.

Page1 / 7

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

This preview shows document pages 1 - 7. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online