Change of entropy

• Pouyan
In summary: I first calculate with 1 kg cold water I get the same answer, but when I add1 kg cold water further, I do not get the same response as the case first. I know the new 1 kg cold water will mix with 1 kg cold & 1 kg warm water which are already mixed, but I don't understand why the entropy change would be different.

Homework Statement

to illustrate an aspect of the second law of thermodynamics, we regard entropy changes in two cases.
We have a vessel with 1 kg of warm water with T = 80 C and another vessel with 2 kg of cold water with T = 10 C and we mix them.

Case 1:
we pour hot water in the vessel with the cold water. The total entropy change becomes S1

Case 2:
we have a third empty vessel. We mix warm water with 1 kg of the cold water in the vessel. The contents of the third container is then poured with the remainder (1 kg) of cold water. The total entropy change becomes S2

calculate S2 / S1

Homework Equations

ΔS = m*c*dT/T
ΔS(tot)= ΔS(hot water)+ΔS(cold water)
c=4190 [J/kg K]
T (hot water) = 80 C = 353 K
T (cold water)= 10 C = 283 K

The Attempt at a Solution

The answer is 1 that means S2 = S1 but I don't get it! I have problem with Case 2

My solution for Case 1 is :

ΔS(hot) = 1 * 4190 * ln(283/353)= -926 J/K
ΔS(cold)=1*4190 * (353-283)/283 = 1036.4 J/K
ΔS1 = 110,4 J/KCase 2 :
We say that we mix first 1kg warm water with 1 kg cold water:
ΔS (hot) = 1 * 4190 * ln(283/353)= -926 J/K (same !)
ΔS (cold) = (1*4190 * (353-283)/283) = 1036.4 J/K (same !)

Then I will mix the other cold water what I think that when we mixed first 1kg of warm water with cold water, we got a new temperature, when we think of Q (missing) = Q (gain)
when the new value is 45 C, which is 318 K.
Now we have 2 kg of water with temperature 318 K in a third vessel and we want to pour 1 kg of water with temperature 283 K. but when I calculate I get it:
ΔS(remainder cold) = 1 * 4190 * ln(318/283) = 488 J/K !

what is the problem ?! how could I think ?! As I said the solution is S2/S1 = 1 ! But I get a bigger S2

Pouyan said:
1
ΔS(hot) = 1 * 4190 * ln(283/353)= -926 J/K
ΔS(cold)=1*4190 * (353-283)/283 = 1036.4 J/K
What is the final temperature of the water?

DrClaude said:
What is the final temperature of the water?

In case 2:
When we first mix 1kg warm and 1 kg cold water
Q(warm water) = Q (cold water)
1 * 4190 (T(warm) - T(average)) = 1* 4190 (T(average)- T(cold))
T average = 45 C

And for case 1? You don't seem to have taken the final temperature into account anywhere.

Pouyan
DrClaude said:
And for case 1? You don't seem to have taken the final temperature into account anywhere.

Should I do that for case 1 too ?!
Ok
we mix 1 kg hot water with 2 kg cold water

1 * 4190 * (80 - T(av)) = 2* 4190 *(T(av)-10)
T(av) = 100/ 3 C= 33.33 C

This is the final temperature in case 1...

Pouyan said:
Should I do that for case 1 too ?!
How can you not do it? To calculate the change in entropy, you need to know the change in temperature.

DrClaude said:
How can you not do it? To calculate the change in entropy, you need to know the change in temperature.
ok now I think I understand what you mean
the first case I used this formula:
m * 4190 * ln (T2 / T1)

but I could also do so here
S = mcdT / T (average)

T (average) is = (C 80 + C 10) / 2 = 45 C
and we must write it in Kelvin when 45 +273 = 318 K

S (heating water) = 1 * 4190 * (10-80) / 318 = -922 J / K

Was it that you meant?!

Pouyan said:
the first case I used this formula:
m * 4190 * ln (T2 / T1)

but I could also do so here
S = mcdT / T (average)
The first equation for ΔS is correct, not the second one. What is incorrect is the value of T2 you were using.

Pouyan said:
T (average) is = (C 80 + C 10) / 2 = 45 C
This is not the final temperature. You need to take into account the amount of water.

DrClaude said:
The first equation for ΔS is correct, not the second one. What is incorrect is the value of T2 you were using.This is not the final temperature. You need to take into account the amount of water.
OK thanks!
but what is the value of T(2) ?!
And my important question is why there is no difference between case 1 and case 2 ?!

the case 1 is clear but the second case , when I first calculate with 1 kg cold water I get the same answer, but when I add1 kg cold water further, I do not get the same response as the case first. I know the new 1 kg cold water will mix with 1 kg cold & 1 kg warm water which are already mixed, that made me dizzy!

please can somebody help me ?

If s0 is the entropy of 1 kg of water at 273 K, then the entropy of 1 kg of water at temperature T is:
$$s=s_0+c\ln{(T/273)}$$
where c is the heat capacity. Does that make sense to you so far?

Chet

Pouyan
Chestermiller said:
If s0 is the entropy of 1 kg of water at 273 K, then the entropy of 1 kg of water at temperature T is:
$$s=s_0+c\ln{(T/273)}$$
where c is the heat capacity. Does that make sense to you so far?

Chet

thanks but why 273 ?! the cold water is 283 k, in case two for the last process, we have a vessel with 2 kg water at 318 K and now we want to pure 1 kg cold water with T = 283 K in that vessel ... This part make me dizzy

Pouyan said:
thanks but why 273 ?! the cold water is 283 k, in case two for the last process, we have a vessel with 2 kg water at 318 K and now we want to pure 1 kg cold water with T = 283 K in that vessel ... This part make me dizzy
I just arbitrarily chose 273 as a reference temperature, with an entropy at this reference temperature of so. This choice will not affect the final answer.

So, I ask again. Are you comfortable with this so far?

Chet

Pouyan
Chestermiller said:
I just arbitrarily chose 273 as a reference temperature, with an entropy at this reference temperature of so. This choice will not affect the final answer.

So, I ask again. Are you comfortable with this so far?

Chet

Yes I think I do ... I know that is the change in entropy...

You don't seem too sure. But OK, here we go.

We are going to focus for now exclusively on Case 1. After we successfully do Case 1, we will analyze Case 2.

CASE 1:

Initial thermodynamic equilibrium state of system:
1 kg water at 80 C (353 K)
2 kg water at 10 C (283 K)Final thermodynamic equilibrium state of system:
3 kg water at temperature ??What is the temperature of the water in the final equilibrium state of the system?

In terms of s0, using the equation in post #11 (which applies to 1 kg of water), what is the entropy of the system in the initial thermodynamic equilibrium state?

In terms of s0, what is the entropy of the system in the final thermodynamic equilibrium state?

Pouyan
Chestermiller said:
You don't seem too sure. But OK, here we go.

We are going to focus for now exclusively on Case 1. After we successfully do Case 1, we will analyze Case 2.

CASE 1:

Initial thermodynamic equilibrium state of system:
1 kg water at 80 C (353 K)
2 kg water at 10 C (283 K)Final thermodynamic equilibrium state of system:
3 kg water at temperature ??What is the temperature of the water in the final equilibrium state of the system?

In terms of s0, using the equation in post #11 (which applies to 1 kg of water), what is the entropy of the system in the initial thermodynamic equilibrium state?

In terms of s0, what is the entropy of the system in the final thermodynamic equilibrium state?
Bingo !

I think I solved the problem !

well for case one when we mix 2 kg cold water with 1 kg hot water we have:
1 (353 -T)=2(T-283)
and T = 306.33 K
S(hot)= 1 * 4190*Ln(306.33/353)= -598.68 J/K
S(cold)= 2*4190*Ln(306.33/283)= 655 J/K
S1=56 J/K

And Case 2
When we mix 1 kg cold and hot water our new T will be 318 K
and
4190*ln(318/353) + 4190 *ln(318/280)= 51,57 J/K

when we add 1 kg cold water again T will be 306
2*4190*ln(306/318) + 1*4190*ln(306/283) = 5.2
S2 = 56 J/K

Now I understand it ! because I'm beginner with entropy and I started with Ice in a big lake problem I thought that I should do the same algorithm but I see the algorithm depends on the situation !
Thanks for help!

Pouyan said:
Bingo !

I think I solved the problem !

well for case one when we mix 2 kg cold water with 1 kg hot water we have:
1 (353 -T)=2(T-283)
and T = 306.33 K
S(hot)= 1 * 4190*Ln(306.33/353)= -598.68 J/K
S(cold)= 2*4190*Ln(306.33/283)= 655 J/K
S1=56 J/K

And Case 2
When we mix 1 kg cold and hot water our new T will be 318 K
and
4190*ln(318/353) + 4190 *ln(318/280)= 51,57 J/K

when we add 1 kg cold water again T will be 306
2*4190*ln(306/318) + 1*4190*ln(306/283) = 5.2
S2 = 56 J/K

Now I understand it ! because I'm beginner with entropy and I started with Ice in a big lake problem I thought that I should do the same algorithm but I see the algorithm depends on the situation !
Thanks for help!
OK. So you have no interest in trying the problem by the alternate approach I was going to suggest. Fine.

Chet

Pouyan
Chestermiller said:
OK. So you have no interest in trying the problem by the alternate approach I was going to suggest. Fine.

Chet
what's the alternative method ?!

Pouyan said:
what's the alternative method ?!

Chet

1. What is entropy?

Entropy is a measure of the randomness or disorder in a system. It is a fundamental concept in thermodynamics and is used to quantify the amount of energy that is unavailable for work in a system.

2. What causes a change in entropy?

A change in entropy can be caused by several factors, including changes in temperature, changes in pressure, chemical reactions, and energy transfer.

3. How is the change of entropy calculated?

The change of entropy (ΔS) is calculated by taking the difference between the entropy of the final state (Sf) and the entropy of the initial state (Si). This can be expressed as ΔS = Sf - Si.

4. What is the relationship between entropy and energy?

Entropy and energy are closely related, as entropy is a measure of the amount of energy that is unavailable for work in a system. In general, as a system becomes more disordered, its entropy increases and the amount of available energy decreases.

5. How does a change in entropy affect a system?

A change in entropy can have various effects on a system, depending on the specific circumstances. In some cases, an increase in entropy can lead to a decrease in the system's energy and stability, while in other cases it can lead to an increase in energy and stability. Overall, changes in entropy can impact the behavior and properties of a system.