How Do You Calculate Efficiency in a Carnot Cycle Involving Argon?

In summary, the conversation discusses a problem involving 0.12 moles of argon undergoing a cycle described in a graph, with specific points and values identified. The goal is to find the efficiency using Q values, but it is unclear what Q represents. The equation used is η= 1- Qout/Qin and there is a Qin value between points A and B.
  • #1
stonecoldgen
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Homework Statement


its part VI) of a big problem, where:

0.12 moles of argon (40) undergoes the cycle abca described in the graph. Process ab lies on a 540k isotherm.

In the previious 5 sub problems, i figured that:

Tc=270k
Qbc=670J (lost)
Qca=400J (added)
Wab=370J
Wabca=100J


The PV diagram has points a,b and c where:

a=(1.5X10-3, 3.6X105)
b=(3X10-3, 1.8X105)
c=(1.5X10-3, 1.8X105)







Homework Equations



Q=mc[itex]\Delta[/itex]T

[itex]\Delta[/itex]U=Q+W

[itex]\epsilon[/itex]=1-Qout/Qin=1-Tlow/Thigh

The Attempt at a Solution


So it's basically using Qabca to find the efficiecy. The thing is that I am not sure what to use for the Qs.
 
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  • #2
What is Qs supposed to represent? I don't see it in your problem statement.
 
  • #3
RTW69 said:
What is Qs supposed to represent? I don't see it in your problem statement.

it's a plural for Q, I am not sure what to use for Qin and Qout
 
  • #4
Remember this is NOT a carnot cycle. η= 1- Qout/Qin There is also a Qin between A and B. Do you know what it is? It is the Qout/ the Total Qin.
 
  • #5


First, let's review the Carnot Cycle. This is a theoretical cycle that describes the most efficient way to convert heat into work. It consists of four processes: two isothermal processes and two adiabatic processes.

In your problem, you are given the temperature and moles of argon for one of the isothermal processes (ab), as well as the heat and work values for the entire cycle. To find the efficiency, we need to calculate the heat values for the other processes (bc and ca).

To do this, we can use the equation Q=mcΔT, where Q is the heat, m is the mass, and ΔT is the change in temperature. Since we are dealing with a gas, we can use the ideal gas law (PV=nRT) to find the change in temperature for each process.

For process bc, we know the initial and final volumes, so we can find the change in volume (ΔV) and use that to find the change in temperature. Similarly, for process ca, we know the initial and final pressures, so we can find the change in pressure (ΔP) and use that to find the change in temperature.

Once we have the heat values for all three processes (ab, bc, and ca), we can use the equation ΔU=Q+W to find the change in internal energy for the entire cycle. Since the Carnot Cycle is a closed cycle, the change in internal energy is equal to zero. We can then use the equation for efficiency, ε=1-Qout/Qin, to find the efficiency of the cycle.

I hope this helps with your homework problem. Remember to always check your units and make sure they are consistent throughout your calculations. Good luck!
 

1. What is the Carnot Cycle?

The Carnot Cycle is a theoretical thermodynamic cycle that describes the most efficient way to convert heat into work. It consists of four reversible processes: isothermal expansion, adiabatic expansion, isothermal compression, and adiabatic compression.

2. How does the Carnot Cycle work?

The Carnot Cycle works by utilizing a heat source, such as a hot reservoir, and a heat sink, such as a cold reservoir. The heat source provides heat to the system, which then undergoes expansion and releases some of that heat as work. The remaining heat is then rejected to the heat sink.

3. What is the efficiency of the Carnot Cycle?

The efficiency of the Carnot Cycle is given by the equation e = 1 - (Tc/Th), where Tc is the temperature of the cold reservoir and Th is the temperature of the hot reservoir. This means that the efficiency of the Carnot Cycle is dependent on the temperature difference between the two reservoirs.

4. What are the applications of the Carnot Cycle?

The Carnot Cycle has many practical applications, including refrigeration, heat pumps, and power generation. It is also used as a benchmark for the maximum efficiency that any heat engine can achieve.

5. What are the limitations of the Carnot Cycle?

The Carnot Cycle is a theoretical cycle and cannot be achieved in real-world systems due to the presence of irreversible processes. Additionally, it assumes that the system operates at a constant temperature, which is not always the case. It also does not take into account any practical limitations, such as friction and heat loss.

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