# Find the work and heat transferred in a carnot cycle.

• Kelsi_Jade
In summary, the conversation discusses a problem involving a Carnot cycle with an ideal gas performing an isothermal transformation between temperatures of 400K and 300K. The initial and final volumes are given and the goal is to find the work performed during a cycle and the amount of heat exchanged with the two sources. The conversation discusses finding Qh and Qc, as well as the efficiency of the Carnot cycle and the work for the entire cycle. The values for Qh and Qc are found to be fairly high due to the gas being compressed and able to do a lot of work through heat flow.
Kelsi_Jade
I'm having some difficulty understanding how to find the work and heat transferred in a carnot cycle.

Problem:
One mole of ideal gas (cV=1.5 R) perform a Carnot cycle between the temperature 400 K and 300 K. On the upper isothermal transformation, the initial volume is 1 liter and the final volume is 5 liter. Find the work performed during a cycle and the amount of heat exchanged with the two sources.

I know that since the process is isothermal the temperature remains constant: ΔT=0 but Q≠0.
Vi=1L
Vf=5L
Th=400k
Tc=300k
Qh=Th(Sb-Sa) - where Sb is the maximum system entropy and Sa is the minimum system entropy
Qc=Tc(Sb-Sa)
W=$\int$PdV=(Th-Tc)(Sb-Sa)

My confusion is where I need to find the entropy to solve for the heat exchanged bt the sources is Q is needed to find entropy??

If someone could let me know if I'm going in the right direction equation-wise, or help explain where to go next it would be appreciated.

Hello, Kelsi_Jade. Welcome to PF!

For the isothermal expansion at 400K, you should be able to deduce values of the change in internal energy, ΔU, and and the work done by the gas, W. Then, apply the first law to this expansion.

Kelsi_Jade said:
I'm having some difficulty understanding how to find the work and heat transferred in a carnot cycle.

Problem:
One mole of ideal gas (cV=1.5 R) perform a Carnot cycle between the temperature 400 K and 300 K. On the upper isothermal transformation, the initial volume is 1 liter and the final volume is 5 liter. Find the work performed during a cycle and the amount of heat exchanged with the two sources.

I know that since the process is isothermal the temperature remains constant: ΔT=0 but Q≠0.
Vi=1L
Vf=5L
Th=400k
Tc=300k
Qh=Th(Sb-Sa) - where Sb is the maximum system entropy and Sa is the minimum system entropy
Qc=Tc(Sb-Sa)
W=$\int$PdV=(Th-Tc)(Sb-Sa)

My confusion is where I need to find the entropy to solve for the heat exchanged bt the sources is Q is needed to find entropy??

You just need to find Qh and Qc.

Can you express Qh in terms of Vf and Vi? (hint: how is dQ related to PdV? hint2: how is P related to V during the isothermal process?).

Once you know Qh you can find Qc from Th and Tc since ΔScycle = 0

AM

TSny said:
For the isothermal expansion at 400K, you should be able to deduce values of the change in internal energy, ΔU, and and the work done by the gas, W.

For an isothermal process, isn't the ΔU=0? Then the work, W, would be W=Q?
(rearranged from ΔU= Q-W)

Andrew Mason said:
You just need to find Qh and Qc.
Can you express Qh in terms of Vf and Vi? (hint: how is dQ related to PdV? hint2: how is P related to V during the isothermal process?).
Once you know Qh you can find Qc from Th and Tc since ΔScycle = 0

Well, P=(nRT)/V
and I know W=nRTln(Vf/Vi) and W=Q so, Q=nRTln(Vf/Vi) ?

Kelsi_Jade said:
For an isothermal process, isn't the ΔU=0? Then the work, W, would be W=Q?
(rearranged from ΔU= Q-W)
Well, P=(nRT)/V
and I know W=nRTln(Vf/Vi) and W=Q so, Q=nRTln(Vf/Vi) ?

Yes. That's right. So, you have Qh. Can you find the efficiency of the Carnot cycle from the two temperatures? If so, then you should be able to use the efficiency to find the work for the entire cycle and also QC.

Kelsi_Jade said:
For an isothermal process, isn't the ΔU=0? Then the work, W, would be W=Q?
(rearranged from ΔU= Q-W)

Well, P=(nRT)/V
and I know W=nRTln(Vf/Vi) and W=Q so, Q=nRTln(Vf/Vi) ?

Yes. Write out the expression for ΔS for the isothermal expansion and for the other three parts of the cycle. What do all those ΔS terms add up to? That will enable you to find Qc.

AM

TSny said:
So, you have Qh. Can you find the efficiency of the Carnot cycle from the two temperatures? If so, then you should be able to use the efficiency to find the work for the entire cycle and also QC.
By using Qh=nRTh*ln(Vf/Vi) , Qh=5,349.77J?

If efficiency=W/Qh and...
e=(Th-Tc)/Th

Then,

e=W/Qh=(Th-Tc)/Th

W=[(Th-Tc)/Th]Qh
W=(5,349.77J) [(400K-300K/400K)]=1,337.44J

Andrew Mason said:
Yes. Write out the expression for ΔS for the isothermal expansion and for the other three parts of the cycle. What do all those ΔS terms add up to? That will enable you to find Qc.

ΔScycle=0
ΔS=Qh/Th + Qc/Tc = 0?
So ,

Qc=(-Qh/Th)Tc

Qc=(-5,349.77J/400K)*300K= -4,012.33J
It makes sense for Qc to be (-) because it is heat that is "leaving" the system and going into the reservoir!
Is it expected that the values be so high, though?

Last edited:
Kelsi_Jade said:
By using Qh=nRTh*ln(Vf/Vi) , Qh=5,349.77J?

If efficiency=W/Qh and...
e=(Th-Tc)/Th

Then,

e=W/Qh=(Th-Tc)/Th

W=[(Th-Tc)/Th]Qh
W=(5,349.77J) [(400K-300K/400K)]=1,337.44J
ΔScycle=0
ΔS=Qh/Th + Qc/Tc = 0?
So ,

Qc=(-Qh/Th)Tc

Qc=(-5,349.77J/400K)*300K= -4,012.33J
It makes sense for Qc to be (-) because it is heat that is "leaving" the system and going into the reservoir!
Is it expected that the values be so high, though?
Looks right. At standard temperature and pressure, one mole will occupy 22.4 l. So this is a fairly compressed gas - high pressure. So it will do a lot of work in expanding 5 times. The ability to do this work comes from the heat flow.

AM

1 person

## 1. What is a Carnot cycle?

A Carnot cycle is a theoretical thermodynamic cycle that describes the most efficient way to convert heat into work or vice versa. It consists of two isothermal processes and two adiabatic processes.

## 2. How is work transferred in a Carnot cycle?

In a Carnot cycle, work is transferred during the two adiabatic processes. During these processes, the system expands or compresses without any heat exchange, resulting in a change in internal energy and the transfer of work.

## 3. How is heat transferred in a Carnot cycle?

Heat is transferred during the two isothermal processes in a Carnot cycle. During these processes, the system is in thermal equilibrium with its surroundings and heat is exchanged to maintain a constant temperature.

## 4. What is the efficiency of a Carnot cycle?

The efficiency of a Carnot cycle is the ratio of the work output to the heat input. It is given by the formula: efficiency = (Thot - Tcold) / Thot, where Thot is the temperature of the hot reservoir and Tcold is the temperature of the cold reservoir.

## 5. How does a Carnot cycle differ from other thermodynamic cycles?

A Carnot cycle is a theoretical ideal that cannot be achieved in real systems. It differs from other thermodynamic cycles in that it is reversible, meaning that it can be run in both directions with the same efficiency. It also operates between two fixed temperatures, unlike other cycles that have varying temperatures throughout the process.

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