Thermodynamics - Carnot engine

[ronny45]

Homework Statement

Consider 2 identical bodies of constant heat capacity Cp and with negligble termal expansion coefficients. Initial temperatures are T1 and T2.
a) When they are placed in thermal contact in an adiabatic enclosure, what is their final temperature?
b) The 2 bodies are brought to thermal equilibrium by a Carnot engine operating between them. The size of the cycle is small so that the temperatures of the bodies do not change appreciably during one cycle; thus the bodies behave as resevoirs during one cycle. Show that the final temperature is sqrt(T1T2). Hint: What is the entropy change of the universe for this second process?
(Assume isobaric conditions throughout, and T2>T1).

The Attempt at a Solution

a) seems fairly straight-foward... if the enclosure is adiabatic, there's no heat exchange between the system and the surroundings, so the final temperature = (T1+T2)/2.

b) I really don't know where to start on this one, I've been at it over a week and I'm none the wiser. The hint is only confusing me because i was under the impression that entropy, as a state function, would not change over one cycle. How does this help me?

 

[anathema]

Hello, thank you for your post. I would like to provide some insight and guidance on the questions you have posed.

a) For the first question, you are correct in stating that the final temperature in an adiabatic enclosure would be (T1+T2)/2. This is known as the principle of thermal equilibrium, which states that when two bodies of different temperatures are in contact, heat will flow from the hotter body to the colder one until they reach the same temperature. In this case, since there is no heat exchange with the surroundings, the final temperature will be the average of the initial temperatures.

b) The second question involves the concept of a Carnot engine, which is a theoretical engine that operates between two thermal reservoirs and is the most efficient heat engine possible. In this case, the two bodies of different temperatures are acting as the thermal reservoirs. The key to solving this problem lies in understanding the Carnot cycle and the principles of thermodynamics.

In a Carnot cycle, the engine operates in a reversible manner, meaning that it can be easily reversed without any loss of energy. This also means that the entropy change of the universe is zero. In this case, the entropy change of the universe can be calculated as the sum of the entropy changes of the two bodies. Since the system is assumed to be isobaric (constant pressure), the entropy change can be calculated as ΔS = Cp ln(T2/T1). This equation can be derived from the definition of entropy, which is ΔS = Q/T, where Q is the heat transferred and T is the temperature.

Now, if we set the entropy change of the universe to be zero, we can solve for the final temperature T3, which is the same for both bodies. This gives us the equation T3 = sqrt(T1T2). This means that the final temperature of the two bodies in thermal equilibrium after one Carnot cycle is the geometric mean of the initial temperatures. This result is in accordance with the second law of thermodynamics, which states that heat cannot spontaneously flow from a colder body to a hotter one.

I hope this explanation helps you in understanding the problem and how to approach it. Remember, in thermodynamics, it is important to understand the concepts and principles rather than just memorizing equations. Good luck with your studies!

 

[thomate1]

I would like to start by clarifying the concept of a Carnot engine. A Carnot engine is a theoretical engine that operates between two heat reservoirs at different temperatures and produces the maximum possible work output. It is based on the principles of thermodynamics and is often used as a benchmark to compare the efficiency of real engines.

In response to the first part of the question, the final temperature of the two bodies in an adiabatic enclosure would indeed be (T1+T2)/2, as there is no heat exchange between the bodies and the surroundings. This is a simple application of the first law of thermodynamics, which states that the change in internal energy of a system is equal to the heat added to the system minus the work done by the system.

Moving on to the second part of the question, we are asked to show that the final temperature of the two bodies brought to thermal equilibrium by a Carnot engine is sqrt(T1T2). To do this, we need to understand the concept of entropy change in a thermodynamic process.

Entropy is a measure of the disorder or randomness in a system. In a reversible process, the change in entropy of the universe (system + surroundings) is zero, as the system and surroundings can be returned to their initial states without any net change in entropy. However, in an irreversible process, the change in entropy of the universe is always positive, as some energy is dissipated as heat and cannot be recovered.

In the case of the Carnot engine, the process is reversible and therefore the change in entropy of the universe is zero. This means that the change in entropy of the system is equal to the change in entropy of the surroundings. Since the surroundings are considered to be large heat reservoirs, the entropy change of the surroundings can be approximated as zero. This means that the change in entropy of the system is also zero.

Using the second law of thermodynamics, we can write:

ΔS = Q/T

Where ΔS is the change in entropy, Q is the heat added to the system, and T is the temperature. Since the change in entropy of the system is zero, we can write:

Q1/T1 + Q2/T2 = 0

Since the Carnot engine operates between two reservoirs, the heat added to one reservoir (Q1) is equal to the heat removed from the other reservoir (Q2). This means that Q1 = -