Changes in entropy of an isolated system

[CAF123]

Homework Statement

Consider two fixed volume bricks of mass m1=2kg and m2=1kg with initial temperatures T1=400K and T2=100K. They are enclosed in a system thermally isolated from the surroundings and are made from a material with a heat capacity c_v = 1kJ/kg/K.

A) In Process 1, the bricks are brought into thermal contact. What is their final temperature once they reached thermal equilibrium with each other?
B)In Process 2, the initial conditions are the same, but the maximum amount of work is extracted from the system (e.g via a heat engine). What is the final temperature in this case?
C)Find the change in entropy of processes 1 and 2.
D)Find the amount of work extracted in Process 2.

Homework Equations

Q = c_vmΔT
ΔS ≥ 0 for an isolated system.

The Attempt at a Solution

A) is fine, the final temperature should be between 100 and 400K. I get 300K which is reasonable.

B)If the max amount of work is extracted, then the heat gained by brick 2, Q₂= Q₁ - W. I don't see how to proceed here unless I write an explicit expression for W (but that comes in D)). I did find an expression though, but when I use it, it yields a negative temperature.

C)Entropy lost by brick 1 = entropy gained by brick 2 since system thermally isolated. Hence net change in entropy in process A is zero.

D)For a thermally isolated system, ΔS ≥ 0. The maximum amount of work is extracted from a Carnot engine, which is reversible, so the equality applies. Therefore, ΔS₂-ΔS₁ = 0 = Q₂/T₂-Q₁/T₁=0 = (Q₁-W)/T₂ - Q₁/T₁=0. Rearrange for W gives Q₁(1-T₂/T₁). Is that correct?

Many thanks for any guidance.

 

[voko]

In B), max energy is extracted when the process is reversible, i.e., total change in entropy is zero. You need to write the equations for the entropy change of each brick, equate their sum with zero, and find the final temperature from that.

In C), obviously process 1 is not reversible, so it cannot have the total change in entropy at zero. The method from B) could be adapted here.

In D), you are not using the correct equations. By definition, $T = dQ / dS$. Note the differentials.

 

[CAF123]

Hi voko,

In B), max energy is extracted when the process is reversible

Why is that so?

You need to write the equations for the entropy change of each brick, equate their sum with zero, and find the final temperature from that.

ΔS₂ = Q₂/(T_f-T₂), where Q₂ is the heat that enters brick 2. Similarly, ΔS₁ = -Q₁/(T_f-T₁), where -Q₁ is the heat lost by brick 1. I can also write Q₂ = c_vm₂(T_f-T₂) and Q₁=c_vm₁(T_f-T₁)

 

[voko]

Hi voko,

Why is that so?

This is what Carnot's theorem is about.

ΔS₂ = Q₂/(T_f-T₂), where Q₂ is the heat that enters brick 2. Similarly, ΔS₁ = -Q₁/(T_f-T₁), where -Q₁ is the heat lost by brick 1. I can also write Q₂ = c_vm₂(T_f-T₂) and Q₁=c_vm₁(T_f-T₁)

Same mistake as before. You are using finite quantities, while you should be using differentials.

 

[CAF123]

Same mistake as before. You are using finite quantities, while you should be using differentials.

So like $dS_1 = dq_1/dT_1$ and $dS_2 = -dq_2/dT_2$. Then sum and equate to zero for a reversible process => $dq_1/dT_1 = dq_2/dT_2 = (c_v m_1 dT_1)/dT_1 = (c_v m_2 dT_2)/dT_2$. It looks like I can cancel dT's, but this cannot be right because I end up with a contradiction.

 

[voko]

So like $dS_1 = dq_1/dT_1$ and $dS_2 = -dq_2/dT_2$.

Not sure why, but you are making the whole thng a lot more complex than it needs to be by inventing equations :) The correct formula is $T = dQ/dS$, as I said in #2. Notice the placement of $d$'s and the minus signs.

 

[CAF123]

Not sure why, but you are making the whole thng a lot more complex than it needs to be by inventing equations :) The correct formula is $T = dQ/dS$, as I said in #2. Notice the placement of $d$'s and the minus signs.

I think I see what you mean. T here is the temperature of the external source of heat. So, the entropy change of brick 2, $\Delta S_2 = c_v m_2 \ln \left(T_f/T_2\right)$ and for brick 1, $\Delta S_1 = -c_v m_1 \ln(T_f/T_1)$ after integration. When I sum these and equate to zero, I end up with $T_f = 1600K.$ Is that right? How can I tell?

 

[voko]

I think I see what you mean. T here is the temperature of the external source of heat. So, the entropy change of brick 2, $\Delta S_2 = c_v m_2 \ln \left(T_f/T_2\right)$ and for brick 1, $\Delta S_1 = -c_v m_1 \ln(T_f/T_1)$ after integration.

You are still imaginative about your equations :) Where does that minus sign come from?

When I sum these and equate to zero, I end up with $T_f = 1600K.$ Is that right? How can I tell?

This is manifestly wrong. The temperature is higher than the bricks had originally.

 

[CAF123]

You are still imaginative about your equations :) Where does that minus sign come from?

$dS_2 = dQ_2/T = c_vm_2dT/T \Rightarrow \Delta S_2 = c_v m_2 \ln (T_f/T_2)$. Similarly for brick 1, and the negative is there for brick 1 because it's entropy decreases since heat flows from it to brick 2.

 

[voko]

Thermodynamic temperature is defined by $T = dQ / dS$. The definition is not dependent on the hand-waving of the kind "the heat flows from X to Y". $Q$ is the heat of the body in question, $S$ is the entropy of the body as well. The definition is independent of the state of other bodies, and, as a useful definition, it had better be.

 

[D H]

When I sum these and equate to zero, I end up with $T_f = 1600K.$ Is that right? How can I tell?

You should immediately know that that is wrong, without asking! The only way the final temperature can be greater than that of the hotter brick is if work is performed on the bricks. Here we're using the temperature difference to extract work from the bricks.

$dS_2 = dQ_2/T = c_vm_2dT/T \Rightarrow \Delta S_2 = c_v m_2 \ln (T_f/T_2)$. Similarly for brick 1, and the negative is there for brick 1 because it's entropy decreases since heat flows from it to brick 2.

No! You are over complicating things and making up equations. Don't do that! If $T_f < T_i$, then $\ln ( T_f/T_i)$ will be negative. There's no reason for you to add a negative sign. All that adding that negative sign accomplished for you was to invalidate the expression.

 

[CAF123]

No! You are over complicating things and making up equations.
Don't do that! If $T_f < T_i$, then $\ln ( T_f/T_i)$ will be negative. There's no reason for you to add a negative sign. All that adding that negative sign accomplished for you was to invalidate the expression.

The fact that the entropy of brick 1 decreases is already encoded in the reasons you gave. With this, I end up with $T_f \approx 252 K$. This seems more reasonable, yes? Energy is extracted from the system, so it is sensible for it to be lower than the 300K obtained in A).

 

[Chestermiller]

Hi CAF123. What DH and Voko are saying is "lose the minus sign in the expression for ΔS₁." Then you'll be OK.

 

[CAF123]

Hi Chestermiller,

Hi CAF123. What DH and Voko are saying is "lose the minus sign in the expression for ΔS₁." Then you'll be OK.

That's what I did and I end up with about 252K. I do not see, however, why what I did qualified as 'making up equations'.

 

[D H]

With this, I end up with $T_f \approx 252 K$. This seems more reasonable, yes?

Very reasonable, and also correct. Very good.

 

[CAF123]

Very reasonable, and also correct. Very good.

Thank you D H. For maximum amount of work to be extracted, the engine is operating at Carnot efficiency, and this is a reversible system. For reversible processes, the change in entropy of the system is zero. Hence the change in entropy of process B is 0.

In process A, I calculated that the change in entropy was about 0.8 kJ K^-1, since ΔS₂ = ln(3) kJ K^-1 and ΔS₁ = -0.288 kJ K^-1. For an isolated system, ΔS_system + ΔS_surr ≥ 0. The second term is zero, since the system is isolated, so the change in entropy of the universe is due to the changes in entropy of the system.

The work extracted, by the first law is W = Q₁-Q₂. Q₁ is the heat flowing from brick 1, which (I think) I may write as Q₁ = -c_vm₁(T_f-T₁) and Q₂ = c_vm₂(T_f-T₂). This gives me W approx. 448kJ. Is it okay?

 

[Chestermiller]

Thank you D H. For maximum amount of work to be extracted, the engine is operating at Carnot efficiency, and this is a reversible system. For reversible processes, the change in entropy of the system is zero. Hence the change in entropy of process B is 0.

In process A, I calculated that the change in entropy was about 0.8 kJ K^-1, since ΔS₂ = ln(3) kJ K^-1 and ΔS₁ = -0.288 kJ K^-1. For an isolated system, ΔS_system + ΔS_surr ≥ 0. The second term is zero, since the system is isolated, so the change in entropy of the universe is due to the changes in entropy of the system.

The work extracted, by the first law is W = Q₁-Q₂. Q₁ is the heat flowing from brick 1, which (I think) I may write as Q₁ = -c_vm₁(T_f-T₁) and Q₂ = c_vm₂(T_f-T₂). This gives me W approx. 448kJ. Is it okay?

I don't obtain these results. For ΔS₁, you forgot to multiply by the mass. For the work, I get Q₁+Q₂, with Q₁=296 kJ and Q₂=-152 kJ, where these Q's apply to the engine, and not to the masses.

 

[CAF123]

I don't obtain these results. For ΔS₁, you forgot to multiply by the mass. For the work, I get Q₁+Q₂, with Q₁=296 kJ and Q₂=-152 kJ, where these Q's apply to the engine, and not to the masses.

I also made a negative error in my expression for Q₂ when I calculated the temperature difference, I get W = 144J now with this correction.

 

[Chestermiller]

I also made a negative error in my expression for Q₂ when I calculated the temperature difference, I get W = 144J now with this correction.

Yes. That matches what I got.

Chet