596_PartUniversity Physics Solution

596_PartUniversity Physics Solution - The Second Law of...

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon
The Second Law of Thermodynamics 20-11 EVALUATE: For a cycle 0, U Δ= so by UQW Δ =− it must be that net net QW = for a cycle. We can also check that net 0: U net 1 2 2 3 3 1 2180 J 1050 J 1130 J 0 UU U U →→→ Δ + Δ + Δ = = (Carnot), ee < as it must. 20.41. IDENTIFY: , p Vn R T = so pV is constant when T is constant. Use the appropriate expression to calculate Q and W for each process in the cycle. H . W e Q = SET UP: For an ideal diatomic gas, 5 2 V CR = and 7 2 . p = EXECUTE: (a) 3 2.0 10 J. aa pV 3 bb p R T = so p Vp V = says . ab TT = (b) For an isothermal process, 21 ln( / ). QW n R T VV == ab is a compression, with , ba VV < so 0 Q < and heat is rejected. bc is at constant pressure, so . p p C Qn CT R = Δ V Δ is positive, so 0 Q > and heat is absorbed. cd is at constant volume, so . V V C R = Δ p Δ is negative, so 0 Q < and heat is rejected. (c) 3 2.0 10 J 241 K. (1.00)(8.314 J/mol K) a T nR × = 241 K. nR = 3 4.0 10 J 481 K. (1.00)(8.314 J/mol K) cc c T nR × = (d) 3 3 3 0.0050 m ln (1.00 mol)(8.314 J/mol K)(241 K)ln 1.39 10 J. 0.010 m b ab a V R T V ⎛⎞ = × ⎜⎟ ⎝⎠ 3 7 2 (1.00)( )(8.314 J/mol K)(241 K) 7.01 10 J. bc p C T = = × 3 5 2 (1.00)( )(8.314 J/mol K)( 241 K) 5.01 10 J. ca V C T = ⋅− = −× net 610 J. ab bc ca QQQ Q =++= net net 610 J. WQ (e) 3 H 610 J 0.087 8.7% 7.01 10 J W e Q = = × EVALUATE: We can calculate W for each process in the cycle. 3 ab ab 53 3 (4.0 10 Pa)(0.0050 m ) 2.00 10 J. bc Wp V =Δ= × = × 0. ca W = net 610 J, ab bc ca WWW W which does equal net . Q 20.42. (a) IDENTIFY and SET UP: Combine Eqs.(20.13) and (20.2) to eliminate C Q and obtain an expression for H Q in terms of W , C , T and H . T 1.00 J, W = C 268.15 K, T = H 290.15 K T = For the heat pump C 0 Q > and H 0 Q < EXECUTE: CH ; WQ Q =+ combining this with CC HH QT = − gives H 1.00 J 13.2 J 1 / 1 (268.15/290.15) W Q = −− (b) Electrical energy is converted directly into heat, so an electrical energy input of 13.2 J would be required. (c) EVALUATE: From part (a), H . 1/ W Q = H Q decreases as C T decreases. The heat pump is less efficient as the temperature difference through which the heat has to be &pumped± increases. In an engine, heat flows from H T to C T and work is extracted. The engine is more efficient the larger the temperature difference through which the heat flows. 20.43. IDENTIFY: bc = and is equal to the maximum temperature. Use the ideal gas law to calculate . a T Apply the appropriate expression to calculate Q for each process.
Background image of page 1

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

View Full DocumentRight Arrow Icon
Image of page 2
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 07/16/2011 for the course PHY 2053 taught by Professor Buchler during the Spring '06 term at University of Florida.

Page1 / 5

596_PartUniversity Physics Solution - The Second Law of...

This preview shows document pages 1 - 2. Sign up to view the full document.

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