Entropy -- can only calculate entropy only at constant temperature?

Outrageous

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

Andrew Mason

Start with:

[itex]\Delta S = \int_{rev} dq/T[/itex]

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

Outrageous

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?

Andrew Mason

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

Outrageous

Thank you.

The assumed reversible path mean doing something infinitesimally?

Andrew Mason

Yes. You have to use the reversible heat flow, dQ_rev 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

Outrageous

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 ?