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

[zag4life]

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.

(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?

 

[TSny]

Hi, zag4life. Welcome to PF.

How does the work for i $\rightarrow$b compare to the work for i$\rightarrow$b$\rightarrow$f?

 

[zag4life]

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.

 

[TSny]

Good. So, use that to get Q for ib.

 

[Nickie]

For part (d), we can use the First Law of Thermodynamics to solve for Qib:

Eint,b = Qib - Wib
27 cal = Qib - 15 cal
Qib = 42 cal

So, for path ib, Q = 42 cal.

For part (e), we can use the same equation to solve for Qbf:

Eint,b = Qbf - Wbf
27 cal = Qbf - (-19 cal)
Qbf = 8 cal

So, for path bf, Q = 8 cal.