{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

5_60_lecture9

# 5_60_lecture9 - MIT OpenCourseWare http/ocw.mit.edu 5.60...

This preview shows pages 1–3. Sign up to view the full content.

MIT OpenCourseWare http://ocw.mit.edu 5.60 Thermodynamics & Kinetics Spring 2008 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms .

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

View Full Document
5.60 Spring 2008 Lecture #9 page 1 Entropy For a reversible ideal gas Carnot cycle: rev Efficiency ε = w = 1 + q 2 rev = 1 T 2 q rev q 1 T 1 q 1 + q 2 = 0 v đ q rev = 0 T 1 T 2 T The efficiency of any reversible engine has to be the same as the Carnot cycle: ( w ) ( w ) ε = ε ′ = q 1 q 1 w Assume ε > ε (left Carnot engine less efficient cycle than Carnot cycle) T 1 (hot) T 2 (cold) q 2 q 1 w q 2 Some reversible engine Since the engine is reversible, we can run it backwards. Use the work (- w ’) out of the Carnot engine as work input ( w ) to run the left engine backwards. Total work out = 0 (- w ’ = w > 0) But ε ' > ε q w 1 > q w 1 w q 1 > q w 1 = w q 1 q 1 < − q 1 since q 1 < 0, q 1 ′ > 0 ( q 1 ′ + q 1 ) > 0 This contradicts the 2 nd law (Clausius). This says that we have a net flow of heat into the hot reservoir, but no work is being done!
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}