Need help understanding entropy

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In summary: This allows you to calculate the energy unavailable for work in this specific scenario, using the entropy change from the previous question. In summary, the entropy change found in B) can be used in C) because it represents the total entropy change of the universe, and the T0 value can be adjusted to calculate the energy unavailable for work in different scenarios.
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maos
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Hey. This isn't exactly homework help. I'm studying for a test and i ran into this:

A) Find the equilibrium temperature that results when 1.8 kilograms of liquid water at 380 K is added to 3.0 kilograms of liquid water at 280 K in a perfectly insulated container.

I answer 318, no problem

B) When heat is added to or removed from a solid or liquid of mass m and specific heat capacity c, the change in entropy can be shown to be S = mc ln(Tf/Ti), where Ti and Tf are the initial and final Kelvin temperatures. Use this equation to calculate the entropy change for each amount of water. Then combine the two entropy changes algebraically to obtain the total entropy change of the universe. Note that the process is irreversible, so the total entropy change of the universe is greater than zero.

I answer 224.4, no problem

Now here is where the problem starts:

C) Assuming that the coldest reservoir at hand has a temperature of 273 K, determine the amount of energy that becomes unavailable for doing work because of the irreversible process.

I used 273*224.4 because Energy unavailable for doing work = T0∆S. and i get the right answer. But why can i use the change in entropy i found in the previous question?

The entropy found in B) came from completely different temperature changes. Why is it that i can use that entropy in C) where the cold reservoir is now 273K?

I don't know if I'm really making sense, :( this unit's been a mess for me.
 
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The entropy change you found in B) is the total entropy change of the universe. This means that it takes into account all energy transfers, regardless of the temperature of the reservoir. In C), you are simply using the same entropy change equation, but with different values for T0 and ∆S. The value of ∆S is the same because it is the total entropy change of the universe, but the T0 value is different because it is now 273 K (the coldest reservoir temperature).
 
  • #3


Hello there,

I understand your confusion with the concept of entropy and its application in different scenarios. Entropy is a measure of the disorder or randomness in a system, and it always increases in an irreversible process. In this case, the mixing of two different temperatures of water is an irreversible process, and as a result, the total entropy change of the universe will be greater than zero.

In part B), you calculated the entropy change for each amount of water separately using the given equation. However, in part C), you are asked to determine the amount of energy that becomes unavailable for doing work because of the irreversible process. This energy is related to the total entropy change of the universe, not just the entropy change of each amount of water separately.

The reason why you can use the entropy change found in part B) in part C) is because the total entropy change of the universe is a state function. This means that it only depends on the initial and final states of the system, not the path taken to reach those states. In this case, the initial and final states of the system are the same in both part B) and part C), so the total entropy change of the universe will be the same. Therefore, you can use the entropy change found in part B) to calculate the energy unavailable for doing work in part C).

I hope this helps to clarify your understanding of entropy. Keep studying and good luck on your test!
 

1. What is entropy?

Entropy is a scientific concept that measures the amount of disorder or randomness in a system. It is often described as a measure of the amount of energy that is no longer available for useful work.

2. How is entropy related to the Second Law of Thermodynamics?

The Second Law of Thermodynamics states that in a closed system, the amount of disorder or randomness (entropy) will tend to increase over time. This means that energy will naturally disperse and become less organized, resulting in an increase in entropy.

3. What is the difference between entropy and enthalpy?

Enthalpy is a measure of the total energy of a system, including both its internal energy and the energy required to create or maintain its structure. Entropy, on the other hand, only considers the disorder or randomness within a system.

4. How can entropy be calculated or measured?

Entropy can be calculated using mathematical equations, such as the Boltzmann equation, based on the number of possible microstates in a system. It can also be measured experimentally using techniques such as calorimetry or statistical mechanics.

5. What are some real-world examples of entropy?

Examples of entropy in everyday life include the spreading of a spilled drink, the rusting of metal, the decay of radioactive materials, and the dispersion of heat in a room. In all of these cases, energy is becoming more dispersed and less organized, resulting in an increase in entropy.

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