Carnot Cycle: Exploring Heat Efficiency and Adiabatic Processes

In summary, the conversation discusses the concept of a Carnot cycle and its complexity, as well as the idea of designing an engine with maximum efficiency by having a system at a slightly higher temperature than the hot reservoir and continually expanding isothermally. However, this violates the Zeroth Law of thermodynamics and is physically impossible. The efficiency of an engine is given by the formula e = 1- Tc/Th, where a smaller temperature for the cold reservoir leads to a higher efficiency. This is because in order to have maximum efficiency, the low-temperature isotherm leg should not lose any heat, which is only possible when the gas is at T=0.
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aaaa202
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I have some trouble understanding the carnot cycle and why you have to make it so complicated, i.e. this process involves 2 isothermal and 2 adiabatic processes.

If you want an engine with maximum efficiency why can't you just have a system whose temperature is just a tiny inch above your hot resevoir and then absorbs heat at a rate such that it continually expands isothermally? - by this I mean that all the heat goes to producing work and keeping the proces isothermal.

Edit: I also have a second question. The efficiency of an engine is given by:

e = 1- Tc/Th , where Tc and Th are the temperatures of the hot and cold resevoir. You can see that the smaller temperature for the cold resevoir the higher efficiency - why is that? Then the entropy expelled to the environment is bigger but how does that make you lose less energy?
 
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aaaa202 said:
I have some trouble understanding the carnot cycle and why you have to make it so complicated, i.e. this process involves 2 isothermal and 2 adiabatic processes.

If you want an engine with maximum efficiency why can't you just have a system whose temperature is just a tiny inch above your hot reservoir and then absorbs heat at a rate such that it continually expands isothermally? - by this I mean that all the heat goes to producing work and keeping the proces isothermal.
You actually can design something that converts heat to work of this sort: for example, a steam turbine basically causes water to isothermally expand into steam forever. The problem with efficiency here is that in addition to the heat you supply to the system, you also are supplying chemical energy in the form of more water (which has a nonzero chemical potential.) The heat engines that we study in elementary thermodynamics are considered to have a constant number of particles N.

The short answer is that an engine of this sort, with constant N, violates the Zeroth Law of thermodynamics. Basically what you're saying is: put the reservoir and the system, initially at different temperatures, in thermal contact. Now we let them exchange heat. Instead of reaching equillibrium (both temperatures equal), they both stay isothermal--the system, which receives heat, just expands isothermally forever, and they never reach thermal equilibrium. Not only does this violate the Zeroth law (everything comes to equilibrium eventually), but it's physically nonsense--how could anything expand infinitely?

If the engine's substance is an ideal gas (let's say one mole), p=RT/V so the work done going from V1 to V2 is
∫p dV = ∫ RT/V dV = RT ln (V2/V1)
If the gas expands at a constant rate, V2/V1 → 1 as t→∞, so the work done is 0. So in order to do work at a constant rate, the medium must expand exponentially faster and faster. Even more unphysical.
Edit: I also have a second question. The efficiency of an engine is given by:

e = 1- Tc/Th , where Tc and Th are the temperatures of the hot and cold resevoir. You can see that the smaller temperature for the cold resevoir the higher efficiency - why is that? Then the entropy expelled to the environment is bigger but how does that make you lose less energy?

The nicer way to look at this is in terms of heat and work. We define Efficiency η = W/Qh

Qh-Ql=W by conservation of energy.
=> η=(Qh-Ql)/Qh, so Ql must be zero for maximum efficiency. Remember now that Ql happens during an isothermal leg--the substance is already in thermal equilibrium with the Tl reservoir at the end of the adiabatic leg. The heat transferred along an isothermal leg is given by -RT ln (Vi/Vf), and this is zero only when T=0 (since the volume is changing). So you want your low-temperature isotherm leg to not lose any heat Ql, and the only time this is possible in an expansion/compression is when the gas is at T=0, where it has no heat to give off.
 
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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 was developed by French physicist Sadi Carnot in 1824.

2. How does the Carnot Cycle work?

The Carnot Cycle consists of four different processes: isothermal expansion, adiabatic expansion, isothermal compression, and adiabatic compression. These processes involve the transfer of heat and work between a system and its surroundings, and follow a specific sequence to maximize efficiency.

3. What is the importance of the Carnot Cycle?

The Carnot Cycle is important because it provides a theoretical upper limit for the efficiency of any heat engine. This means that no real engine can ever be more efficient than a Carnot engine, making it a benchmark for the study and improvement of real-world engines.

4. What is an adiabatic process in the Carnot Cycle?

An adiabatic process in the Carnot Cycle is one in which there is no heat transfer between the system and its surroundings. This means that the temperature of the system changes without any energy being added or removed through heat transfer. Adiabatic processes play a crucial role in maximizing the efficiency of the Carnot Cycle.

5. How is the efficiency of the Carnot Cycle calculated?

The efficiency of the Carnot Cycle is calculated by taking the ratio of the work output to the heat input. This can be expressed as efficiency = (T1 - T2) / T1, where T1 is the temperature at which heat is added and T2 is the temperature at which heat is rejected. This formula shows that the efficiency of the Carnot Cycle is directly related to the temperature difference between the hot and cold reservoirs, and can never exceed 100%.

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