How to integrate a sum of two entropy differentials?

[zenterix]

Homework Statement

Two blocks of the same metal are of the same size but are at different temperatures $T_1$ and $T_2$. These blocks of metal are brought together and allowed to come to the same temperature.

Relevant Equations

Show that the entropy change is given by

$$\Delta S=C_P\ln{\left [ \frac{(T_1+T_2)^2}{4T_1T_2} \right ]}$$

if $C_P$ is constant.

Here is a way to solve the problem.

Since $dq_1=-dq_2$ then

$$\int_{T_1}^T C_PdT=-\int_{T_2}^T C_PdT\tag{1}$$

$$\implies T=\frac{T_1+T_2}{2}\tag{2}$$

$$dq_1=C_PdT\tag{3}$$

$$dS_1=\frac{dq_1}{T}=\frac{C_P}{T}dT\tag{4}$$

$$\Delta S_1=\int_{T_1}^T\frac{C_P}{T}dT=C_P\ln{\frac{T}{T_1}}\tag{5}$$

$$dq_2=C_PdT\tag{6}$$

$$dS_2=\frac{dq_2}{T}=\frac{C_P}{T}dT\tag{7}$$

$$\Delta S_1=\int_{T_2}^T\frac{C_P}{T}dT=C_P\ln{\frac{T}{T_2}}\tag{8}$$

$$\Delta S_{sys}=\Delta S_1+\Delta S_2=C_P\ln{\frac{T^2}{T_1T_2}}=C_P\ln{\frac{(T_1+T_2)^2}{4T_1T_2}}\tag{9}$$

My question is about solving it a different way, if possible.

Can we write

$$dS_{sys}=dS_1+dS_2=\frac{dq_1}{T_1}+\frac{dq_2}{T_2}\tag{10}$$

$$=\frac{dq_1}{T_1}-\frac{dq_1}{T_2}\tag{11}$$

is this expression correct?

I am really not sure. If it is, I am not sure what it means to integrate the right-hand side. In particular, what would the limits of integration be?

 

[Philip Koeck]

Can we write

$$dS_{sys}=dS_1+dS_2=\frac{dq_1}{T_1}+\frac{dq_2}{T_2}\tag{10}$$

$$=\frac{dq_1}{T_1}-\frac{dq_1}{T_2}\tag{11}$$

is this expression correct?

I am really not sure. If it is, I am not sure what it means to integrate the right-hand side. In particular, what would the limits of integration be?

The 2 blocks don't have constant temperatures.
T₁ and T₂ are only their starting temperatures.
Try again. Then you should see what the limits of integration have to be.

 

[zenterix]

The 2 blocks don't have constant temperatures.
T₁ and T₂ are only their starting temperatures.
Try again. Then you should see what the limits of integration have to be.

I am aware that the temperatures aren't constant.

Indeed, the expression $\frac{dq_1}{T_1}-\frac{dq_1}{T_2}$ isn't correct because I defined $T_1$ and $T_2$ as constant initial temperatures.

In the correct solution I showed, when I write $\int_{T_1}^T\frac{C_P}{T}dT$, the $T$, as far as I understand, is actually the external temperature. That is, it is the temperature of the external "reservoir" for a reversible process.

Am I correct to say that implicit in this problem is the assumption that the heat exchange happens reversibly?

I am really not sure. The fact that the calculated entropy change is positive means this entire process can occur spontaneously.

After all, the differential $dS=\frac{dq}{T}$ is only true for reversible processes.

If I write $\frac{dq_1}{t_1}-\frac{dq_1}{t_2}$ then $t_1$ and $t_2$ denote external temperatures. Would this be correct?

I find it incredible how difficult these concepts are even though they look so simple.

 

[Philip Koeck]

I am aware that the temperatures aren't constant.

Indeed, the expression $\frac{dq_1}{T_1}-\frac{dq_1}{T_2}$ isn't correct because I defined $T_1$ and $T_2$ as constant initial temperatures.

In the correct solution I showed, when I write $\int_{T_1}^T\frac{C_P}{T}dT$, the $T$, as far as I understand, is actually the external temperature. That is, it is the temperature of the external "reservoir" for a reversible process.

Am I correct to say that implicit in this problem is the assumption that the heat exchange happens reversibly?

I am really not sure. The fact that the calculated entropy change is positive means this entire process can occur spontaneously.

After all, the differential $dS=\frac{dq}{T}$ is only true for reversible processes.

If I write $\frac{dq_1}{t_1}-\frac{dq_1}{t_2}$ then $t_1$ and $t_2$ denote external temperatures. Would this be correct?

I find it incredible how difficult these concepts are even though they look so simple.

Yes, you do have to consider a reversible process to calculate ΔS.
You could imagine an environment that has all the temperatures between T₁ and T₂ so that the temperature change can occur reversibly, but that just means that you only need to think of the temperatures of the two blocks anyway.
Your equations 10 and 11 are correct if T₁ stands for the varying temperature of block 1 and T₂ for the varying temperature of block 2. When you integrate you just use the starting and final temperatures of each block as limits, so you need two separate integrals.
I would say you end up with the same as the first solution you give.

 

[Philip Koeck]

Am I correct to say that implicit in this problem is the assumption that the heat exchange happens reversibly?

I am really not sure. The fact that the calculated entropy change is positive means this entire process can occur spontaneously.

After all, the differential $dS=\frac{dq}{T}$ is only true for reversible processes.

If I write $\frac{dq_1}{t_1}-\frac{dq_1}{t_2}$ then $t_1$ and $t_2$ denote external temperatures. Would this be correct?

I find it incredible how difficult these concepts are even though they look so simple.

I agree it's strange.

You need an (imagined) reversible process to calculate the entropy change for each block, but in the end you find that the total entropy change is positive, showing that the process i irreversible.

This only makes sense because entropy is a state function.