Physics - Entropy, 2nd law of thermodynamics

• WoAiMeiMei

  17 Apr 08, 20:35

  ahh im having problems with these 2 remaining questions

  SOLVED : P8 (mistake with scale)
                     P28
  

   

  
  Notes:

  
  
  P32 (Part b) - how do i do this? hint is given below
  How are pressure and volume are related in an adiabatic expansion? Can you determine the type of gas from the value of gamma γ? How much heat is involved in an adiabatic process? In which process, A to B or C to D, is heat absorbed? Lost? Can you get the temperature change in terms of the product of pressure and volume? How is efficiency related to the heat gained and heat lost?

   

  

   

  *expired notes*

   

  P8 - The line is not linear.
  i tried formula delta s = cm ln (Tf / Ti)
  which is 125 = c x 0.376 x ln (390 / 300)
  Correction: it should be 125 = c x 0.376 x ln (390 / 315)

   

  
  P28 - W=Qh-QL, Qh/Th=QL/TL and P=W/t.... what do i do? t=1? i did it and it gives me 388 W again. would both answers a and b be the same?

• WoAiMeiMei

  18 Apr 08, 21:51

  help? :x

• justinfoo

  20 Apr 08, 00:54

  If I'm not wrong,

  P28

  You need to find the entropy, S

  Let Sb = entropy of saturated water at 132 + 273 = 405K and Sa = entropy of saturated steam at 405K

  Look for values of both entropy from steam tables....

  Then use : Qh = Th(Sb - Sa)    and Qc = Tc(Sb - Sa)

• WoAiMeiMei

  20 Apr 08, 19:48

  28 part a:

   

  i used e= w/q(h) to give q(h)....

   

  so the answer is like 2219.49 W

   

  trying to do part b

• WoAiMeiMei

  20 Apr 08, 19:50

  woot! part b = 1831.4915 W

   

  i just did...

   

  405 - 334.2 = 1.2118

  1 - ans = efficiency

   

  388 / efficiency = 1831W

• eagle

  20 Apr 08, 21:19

  i forget carnot engine liao... no time to read up now also... :(