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First Law of Thermodynamics (Eint = Q-W) question. I am only stuck on part (d)

  • Thread starter zag4life
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  • #1
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When a system is taken from state i to state f along path iaf in the figure, Q = 60 cal and W = 10 cal. Along path ibf, Q = 65 cal.
hrw7_18-41.gif

(a) What is W along path ibf? (b) If W = -19 cal for the return path fi, what is Q for this path? (c) If Eint,i = 11 cal, what is Eint,f?(d) If Eint,b = 27 cal what is Q for path ib?(e) For the same value of Eint,b, what is Q for path bf?

Homework Equations


Eint= internal energy of the system; Q= heat; W= work
I have used Eint= Q - W to solve most of the problem.

The Attempt at a Solution


I only am marked wrong on part d which is very frustrating. So far I have:
(a)Wibf = 15 cal because W= Q-E (b) Qfi= -69 cal because Q= -(E)+W (c) Eint,f= 61 cal because Eint,f= Eint+Ei (d) Qib= ? (e) Qbf= 34 because Eint,f =E-Eint,b
Any thoughts?
 
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Answers and Replies

  • #2
TSny
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Hi, zag4life. Welcome to PF.

How does the work for i [itex]\rightarrow[/itex]b compare to the work for i[itex]\rightarrow[/itex]b[itex]\rightarrow[/itex]f?
 
  • #3
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Well, from b to f, because the volume does not change, work is equal to zero. The work for i,b,f would be the same as the work for i,b I think.
 
  • #4
TSny
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Good. So, use that to get Q for ib.
 

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