Entropy - can only calculate entropy only at constant temperature?

• Outrageous
In summary: The simplest reversible path for the surroundings would be an isothermal path. This means that there can be no temperature gradients anywhere. All heat flow occurs reversibly: an infinitessimal change in temperature of the body or surroundings can reverse the direction of heat flow.
Outrageous
Entropy -- can only calculate entropy only at constant temperature?

Is that possible to calculate entropy when dq=+100J , temperature change from 271K to 273K ?
Ice melt until 273K , not an isolated system.

Or I can only calculate entropy only at constant temperature?

Thank you

Outrageous said:
Is that possible to calculate entropy when dq=+100J , temperature change from 271K to 273K ?
Ice melt until 273K , not an isolated system.

Or I can only calculate entropy only at constant temperature?

$\Delta S = \int_{rev} dq/T$

So in order to determine the change in entropy of the ice, you have to break the process into two parts:

first the ice is warmed up from 271K to 273K. What is dq in terms of dT and the mass of the ice? (hint: you have to know the specific heat capacity of ice).

second: the ice melts all at 273K, so the ΔS will be easy to calculate.

Then all you have to do is add up the changes in entropy to get the total change in entropy.

AM

Thank you
But if I want to calculate the entropy of surrounding?
Then I have to use specific heat capacity of ice and latent heat fusion of ice.it will same as the entropy forward, then the change of entropy will be zero.
But I think that is irreversible process... Am I wrong?

Outrageous said:
Thank you
But if I want to calculate the entropy of surrounding?
Then I have to use specific heat capacity of ice and latent heat fusion of ice.it will same as the entropy forward, then the change of entropy will be zero.
But I think that is irreversible process... Am I wrong?
The change in entropy of the surroundings is very easy, since the process is essentially isothermal (the temperature of the surroundings does not change during the process). The reversible path for the surroundings is just an isothermal reversible path in which the heat flowing out of the surroundings and into the ice does so at constant temperature. (1) What is the total heat flow into the surroundings? (2) What is the temperature. Divide (1) by (2) to get ΔS of the surroundings.

When you have an irreversible process, to calculate the entropy of the system and of the surroundings you have to use a different reversible path for each. In the case of the ice going from 271K to 273K, you have to use a quasi-static path in which the ice is in thermal contact with a body at an infinitessimally higher temperature that keeps increasing slowly as the ice absorbs heat flow, Q. For the surroundings you would use a quasi-static path in which the surroundings are in contact with a body at an infinitessimally lower temperture until Q heat flow leaves the surroundings.

AM

Last edited:

Andrew Mason said:
(1) What is the total heat flow into the surroundings? (2) What is the temperature. Divide (1) by (2) to get ΔS of the surroundings.
Thank you.

Andrew Mason said:
When you have an irreversible process, to calculate the entropy of the system and of the surroundings you have to use a different reversible path for each. In the case of the ice going from 271K to 273K, you have to use a quasi-static path in which the ice is in thermal contact with a body at an infinitessimally higher temperature that keeps increasing slowly as the ice absorbs heat flow, Q. For the surroundings you would use a quasi-static path in which the surroundings are in contact with a body at an infinitessimally lower temperture until Q heat flow leaves the surroundings.
AM
The assumed reversible path mean doing something infinitesimally?

Outrageous said:
The assumed reversible path mean doing something infinitesimally?

Yes. You have to use the reversible heat flow, dQrev in order to calculate entropy change. The simplest reversible path for the surroundings would be an isothermal path. This means that there can be no temperature gradients anywhere. All heat flow occurs reversibly: an infinitessimal change in temperature of the body or surroundings can reverse the direction of heat flow.

In an irreversible process where there is a finite temperature difference between the system and surroundings, the body (in this case it is ice) is not at a uniform temperature as the heat flows (until it reaches 273K). There is a finite temperature gradient within the body and in the surroundings as the heat flows so heat flow is not reversible.

On the other hand, the melting of the ice at 273K is essentially reversible since the temperature of the ice and surroundings are effectively at 273K while the ice is melting.

AM

Andrew Mason said:
eAll heat flow occurs reversibly: an infinitessimal change in temperature of the body or surroundings can reverse the direction of heat flow. AM

Reverse the direction of heat flow? heat is always flow from high temperature to low. Reverse here mean we make heat flow from ice to the surrounding, how ?

1. What is entropy and why is it important in science?

Entropy is a measure of the disorder or randomness in a system. It is important in science because it helps us understand the direction of spontaneous processes and the efficiency of energy conversions.

2. Can entropy only be calculated at a constant temperature?

Yes, entropy can only be accurately calculated at a constant temperature. This is because temperature is a key factor that affects the randomness of molecules and their ability to move and change.

3. How do changes in temperature affect entropy?

An increase in temperature leads to an increase in entropy, as molecules have more energy and can move more randomly. Conversely, a decrease in temperature leads to a decrease in entropy as molecules have less energy and move less randomly.

4. Can entropy be negative?

Yes, entropy can be negative. This typically occurs when there is a decrease in disorder or randomness within a system, such as when a solid forms from a liquid. However, in the overall universe, entropy will always increase.

5. How is entropy related to the second law of thermodynamics?

The second law of thermodynamics states that the total entropy of an isolated system will always increase over time. This means that the disorder or randomness within a system will always increase, and energy will always be lost in the form of heat. Entropy is a key concept in understanding this law.

Replies
3
Views
1K
Replies
15
Views
2K
Replies
9
Views
2K
Replies
16
Views
1K
Replies
10
Views
2K
Replies
3
Views
2K
Replies
13
Views
2K
Replies
3
Views
1K
Replies
12
Views
2K
Replies
2
Views
2K