How Do You Solve Thermodynamic Cycle Problems with Missing Values?

[NBAJam100]

Homework Statement

Im given a table with values for Delta(Internal energy), Delta(KE), Delta(PE), Delta(E), Q, and W.

For certain ones, certain values are missing and I have to fill in the blanks.

Specifically- D()= Delta
Process 1-2) D(U) =72, D(KE)= =5 , D(PE)= -6, D(E) = ? , Q=0 , W= ?
Process 2-3.) D(U)=64 , D(KE)=0 , D(PE) = ? , D(E) = ? , Q = 90, W= ?

Homework Equations

Delta(E)= Delta(U) +Delta(KE) + Delta(PE)

Delta(E)=Q-W

The Attempt at a Solution

For Process 1-2, I found the values of D(E) and W because there were only 2 unkowns so I could plug it into the given equations and solve for the unknowns. For Process 2-3 (And the rest of the remaining processes not listed) I have no idea what to do because there are 3 unknowns and 2 equations that I see fit to use. I've tried re-arranging and plugging different values in but I am stuck. Can someone give me a hand?

 

[LightHero]

For Process 1-2: D(E) = 72+5-6 D(E)=71Delta(E)=Q-W 71=0-WW=-71For Process 2-3:D(PE)= D(U)- D(KE) D(PE)=64-0 D(PE)=64 Maybe I could use this to solve for the other equations but I'm still not sure how..

 

[Mene]

It seems like you are on the right track with using the equations Delta(E)= Delta(U) +Delta(KE) + Delta(PE) and Delta(E)=Q-W to solve for the missing values. However, for Process 2-3, you will need to use an additional equation, which is the first law of thermodynamics: Delta(U)=Q-W. This equation states that the change in internal energy is equal to the heat added to the system minus the work done by the system.

Using this equation, you can solve for the missing values of D(PE) and D(E) by setting up a system of equations with the given values and the unknowns. The equations will look like this:

D(U) = 64 = Q - W
D(E) = ? = D(U) + D(KE) + D(PE)
Q = 90
W = ?

From these equations, you can solve for the missing values of D(PE) and D(E) by substituting in the given values and solving for the unknowns.

It is important to note that in thermodynamic cycles, the first law of thermodynamics must always hold true, meaning that the total energy added to the system (Q) must be equal to the work done by the system (W) plus the change in internal energy (D(U)). So, you can also use this fact to check your solutions and ensure that they are correct.

I hope this helps with solving the remaining processes and understanding the concepts behind it. Good luck!