Evaluating a cycle [Thermodynamics]

In summary: So the system would not have to expend any work to expel heat in this case.This is not the same situation. In the first situation, the system is in thermal contact with the environment. In the second situation, the system is isolated from the environment.The system must expend work to expel heat in this case.
  • #1
guyvsdcsniper
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Homework Statement
Cycle. A system undergoes the following four- stage cyclic process:
In stage (1) the system absorbs 226 cal of heat and does 50 J of work; in
(2) the system adiabatically does 30 J of work; and in (3) the system rejects
100 cal of heat while the environment does 80 J of work on the system.
Stage (4) is also adiabatic. Is work done on or by the system in stage (4)?
How many joules of work?
Relevant Equations
Q+W=0
Since this is a cycle, there should leave the system unchanged, hence Q+W = 0.

I wrote evaluated the heat and work done on/by the system at each cycle as shown in my work.

The first question ask is work done at the fourth stage. Since this a cycle and again the system she remain unchanged, and assuming the fourth stage is the last stage I can try to determine this by analyzing each cycle. The first stage absorbs 943.036J of heat. At the third stage, -418.6J of heat is expelled. That makes ΔQ = 527.436.

If the fourth stage is adiabatic, no heat can transfer at this stage. But since this is a cycle, the system must remain unchanged at the end. So that means there must be work done at the fourth stage in order to return back to the overall energy we started with.

We can say ΔQ=-W with W being each work done at each stage, -(W1+W2+W3+W4)

By algebra W4 = -527.436J.

So the system does work at this stage.

Is my approach correct?

Screen Shot 2022-02-05 at 1.32.53 PM.png
 
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  • #2
Your approach is correct and your numbers match mine.
 
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  • #3
kuruman said:
Your approach is correct and your numbers match mine.
The end result should be negative right? I forgot to carry the negative in my original solution and had to add it in the red.
 
  • #4
quittingthecult said:
The end result should be negative right? I forgot to carry the negative in my original solution and had to add it in the red.
You got a negative value for ##W_4##. But now you have to interpret this to decide if 527 J of work was done on or by the system during step 4.
 
  • #5
TSny said:
You got a negative value for ##W_4##. But now you have to interpret this to decide if 527 J of work was done on or by the system during step 4.
It should be done by the system on step 4 correct?
 
  • #6
quittingthecult said:
It should be done by the system on step 4 correct?
Yes, that's right. Can you give the reason for this?
 
  • #7
TSny said:
Yes, that's right. Can you give the reason for this?
So if the fourth stage is adiabatic, this means that no heat can transfer at this stage. And this is also a cycle so the system must remain unchanged at the end. So Q+W=0.

Up until stage 4 all the work done on/by the system has canceled out and we only have a change of heat remaining. The heat was positve and when Q > 0 heat is absorbed by the system. Since W < 0, Work is done by the system in order to expell that internal energy/heat?

Is that correct?
 
  • #8
quittingthecult said:
So if the fourth stage is adiabatic, this means that no heat can transfer at this stage. And this is also a cycle so the system must remain unchanged at the end. So Q+W=0.

Up until stage 4 all the work done on/by the system has canceled out and we only have a change of heat remaining. The heat was positve and when Q > 0 heat is absorbed by the system. Since W < 0, Work is done by the system in order to expell that internal energy/heat?

Is that correct?
Yes. Good.

As you stated in your first post, ##Q+W = 0## for any cycle. This assumes that ##W## is defined as work done on the system. Since you found ##W_4 =## -527 J, it means that -527 J was done on the system in step 4. This is equivalent to saying +527 J was done by the system in step 4.

You were probably already clear on this, but I just wanted to make sure. Signs can sometimes be tricky. You will find some books writing ##Q-W = 0## for a cycle because they define ##W## as work done by the system.
 
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  • #9
I see now that you already dealt with the signs in this thread. :oldsmile:
 
  • #10
quittingthecult said:
Since W < 0, Work is done by the system in order to expell that internal energy/heat?
Are you really saying that the system must do work in order to expel heat? What if I have an ideal gas at temperature ##T_1## in a rigid box and place the box in good thermal contact with a reservoir at a lower temperature ##T_2##? The gas will expel heat without exchanging any work with the environment because its volume does not change. The expelled heat comes entirely at the expense of the internal energy.
 

Related to Evaluating a cycle [Thermodynamics]

1. What is the first law of thermodynamics and how does it relate to evaluating a cycle?

The first law of thermodynamics states that energy cannot be created or destroyed, only transferred or converted from one form to another. This law is important in evaluating a cycle because it helps us understand how energy is conserved and how it is transferred throughout the various stages of a cycle.

2. What is the difference between an open and closed thermodynamic cycle?

An open thermodynamic cycle is one in which matter can enter or leave the system, while a closed cycle is one in which no matter can enter or leave the system. This difference is important in evaluating a cycle because it affects the amount of energy that is transferred and the overall efficiency of the cycle.

3. How do you calculate the efficiency of a thermodynamic cycle?

The efficiency of a thermodynamic cycle can be calculated by dividing the work output by the energy input. This is known as the thermal efficiency and is a measure of how well the cycle converts energy into work. The higher the efficiency, the more efficient the cycle is at converting energy.

4. What are some common types of thermodynamic cycles?

Some common types of thermodynamic cycles include the Carnot cycle, the Rankine cycle, and the Otto cycle. These cycles are used in various applications such as power plants, refrigeration systems, and internal combustion engines. Each cycle has its own unique characteristics and efficiency.

5. How does entropy play a role in evaluating a thermodynamic cycle?

Entropy is a measure of the disorder or randomness of a system. In a thermodynamic cycle, entropy can increase or decrease depending on the type of cycle and the processes involved. It is important to consider entropy when evaluating a cycle because it can affect the overall efficiency and performance of the system.

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