I am stuck on a calculation -- Entropy change for a compound system

[Lambda96]

Homework Statement

show that $$ \Delta S \geqslant 0 $$

Relevant Equations

none

Hi,

Unfortunately, I have problems with the task 4

In task 3 I got the following

$$ T_f=T_ie^{\Delta S_i - c_i} $$

Then I proceeded as follows

$$ \Delta S = \Delta S_1 + \Delta S_1 $$
$$ \Delta S =c_1ln(\frac{T_ie^{\Delta S_i - c_i}}{T_1})+c_2ln(\frac{T_f}{T_2})$$
$$ \Delta S =c_1ln(\frac{T_1e^{\Delta S_1 - c_1}}{T_1})+c_2ln(\frac{T_2e^{\Delta S_2 - c_2}}{T_2})$$
$$ \Delta S = c_1ln(e^{\Delta S_1 - c_1}) + c_2*ln(e^{\Delta S_2 -c_2}) $$
$$ \Delta S = c_1*(\Delta S_1 - c_1) + c_2*(\Delta S_2 - c_2) $$

Unfortunately, I can not get any further now

Did I do anything wrong, or am I missing something that I can show that this expression is greater than or equal to 0?

 

[Philip Koeck]

Homework Statement:: show that $$ \Delta S \geqslant 0 $$
Relevant Equations:: none

Hi,

Unfortunately, I have problems with the task 4View attachment 317365
In task 3 I got the following

$$ T_f=T_ie^{\Delta S_i - c_i} $$

Then I proceeded as follows

$$ \Delta S = \Delta S_1 + \Delta S_1 $$
$$ \Delta S =c_1ln(\frac{T_ie^{\Delta S_i - c_i}}{T_1})+c_2ln(\frac{T_f}{T_2})$$
$$ \Delta S =c_1ln(\frac{T_1e^{\Delta S_1 - c_1}}{T_1})+c_2ln(\frac{T_2e^{\Delta S_2 - c_2}}{T_2})$$
$$ \Delta S = c_1ln(e^{\Delta S_1 - c_1}) + c_2*ln(e^{\Delta S_2 -c_2}) $$
$$ \Delta S = c_1*(\Delta S_1 - c_1) + c_2*(\Delta S_2 - c_2) $$

Unfortunately, I can not get any further now

Did I do anything wrong, or am I missing something that I can show that this expression is greater than or equal to 0?

Stop manipulating expressions and start thinking, I would say.

Your result for task 3 is both mathematically wrong (It doesn't follow from the expression in task 1) and it doesn't help you as long as you don't know ΔS for either of the systems, and for that you need T_f.
You are going round in circles!

So, you need to determine T_f first, without using ΔS.

 

[Lambda96]

Thanks for your help Philip

I thought about the task again. The two systems should have the same temperature $T_f$ at the end. The heat transfer is calculated as follows

$$ \Delta Q = C\Delta T $$

Then the following must be true for the system.

$$\Delta Q_1=-\Delta Q_2$$
$$C_1\Delta T=-C_2\Delta T$$
$$C_1(T_f-T_1)=-C_2(T_f-T_2)$$

I can now solve this expression for the final temperature and get

$$T_f=\frac{C_1T_1+C_2T_2}{C_1+C_2}$$

Back to the task

$$\Delta S=C_1ln(\frac{T_f}{T_1}+C_2ln(\frac{T_f}{T_2})$$

Unfortunately, I'm not getting anywhere now because I can't think of anything to do with the natural logarithm.

 

[kuruman]

You are off the mark. $$\Delta S=\int_{T_i}^{T_f}\frac{dQ}{T}=\int_{T_i}^{T_f}\frac{CdT}{T}=C\ln\left(\frac{T_f}{T_i}\right).$$Thus, for the two systems, $$\Delta S=\Delta S_1+\Delta S_2=C_1\ln\left(\frac{T_f}{T_1}\right)+C_2\ln\left(\frac{T_f}{T_2}\right).$$ Proceed from there.

 

[Philip Koeck]

$$\Delta S=C_1ln(\frac{T_f}{T_1}+C_2ln(\frac{T_f}{T_2})$$

Unfortunately, I'm not getting anywhere now because I can't think of anything to do with the natural logarithm.

Maybe split up the logarithms and summarize so you have only one T_f, then insert the expression for T_f and continue splitting and simplifying. Could work.

 

[Lambda96]

Thank you Philip and kuruman for your help I was now able to show that the entropy change is greater than or equal to zero

 

[Lambda96]

Here is my solution

 

[Philip Koeck]

Here is my solution

View attachment 317564

That would mean that the process is reversible, but if T₁ and T₂ are not the same it should be irreversible and ΔS should be larger than 0.

 

[TSny]

 

[Philip Koeck]

View attachment 317570

Meant for the OP:
The expression before the mistake pointed out above by TSny is symmetric in T₁, C₁ and T₂, C₂ so you can simply decide that T₁ > T₂. I believe that will lead to ΔS > 0.

 

[Lambda96]

Thanks TSny and Philip for your help , I have corrected the error before submission, thanks again for pointing it out