Calculating entropy change associated with change in temperature

In summary: It's fixed now.## In summary, The conversation is about calculating entropy change when there is a temperature change involved. The equation for calculating entropy (ΔS) in a reversible process is given, but the correct answer is obtained by chance because the temperature is not constant and the equation needs to be integrated over the range of temperatures. The correct equation is ΔS = ∫(mC/T)dT. Using this equation, the correct answer for the given problem is 4.79 kJ⋅K-1.
  • #1
Ryaners
50
2
Hi folks,
This is a question about how to calculate entropy change when there is a temperature change involved. I got the correct answer to this, but I don't actually understand why it's correct..!
Any help is much appreciated.

Homework Statement


One mole of liquid bromine is heated from 30 ºC to 50 ºC at constant pressure. Calculate the change in entropy (in kJ⋅K-1) associated to the process:

Br2 (liq, 30 ºC) → Br2 (liq, 50 ºC)

knowing that the heat capacity of Br2 at constant pressure in the range 30-50 ºC is 75 kJ⋅mol-1⋅K-1.

Homework Equations


q = m⋅C⋅ΔT
ΔS = qrev / T (for a reversible process)

The Attempt at a Solution


First I calculated q:
(1 mol)(75 kJ⋅mol-1⋅K-1)(20°K) = 1500 kJ
Then I used the ΔS equation above, with T = 298.15 K:
ΔS = 1500 kJ / 298.15 K = 5 kJ⋅K-1

The thing is, this is the correct answer, but I have no idea why you should know to use 298.15 K as the temperature. It's not given in the question, and I would have thought that the above equation for ΔS could only be used in cases where the temperature remains constant, not like in this example where it increases by 20°.

Can anyone explain why?
 
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  • #2
Ryaners said:
ΔS = 1500 kJ / 298.15 K = 5 kJ⋅K-1

The thing is, this is the correct answer, but I have no idea why you should know to use 298.15 K as the temperature.
That's not the right answer. It might be numerically close to the actual answer, but that's serendipitous.

Since T is not a constant, you have to calculate ΔS by integrating over T:
$$
\Delta S = \int_{T_i}^{T_f} \frac{m C}{T} dT
$$
 
Last edited:
  • #3
DrClaude said:
That's not the right answer. It might be numerically close to the actual answer, but that's serendipitous.

Since T is not a constant, you have to calculate ΔS by integrating over T:
$$
\Delta S = \int_{T_i}^{T_f} \frac{m C}{T} dt
$$

Ok, I figured there was something fishy going on. I'll try the integration approach - there was another similar question that I also got the right answer for, but I guess they were both right by chance!

Thanks for your help :)
 
  • #4
DrClaude said:
That's not the right answer. It might be numerically close to the actual answer, but that's serendipitous.

Since T is not a constant, you have to calculate ΔS by integrating over T:
$$
\Delta S = \int_{T_i}^{T_f} \frac{m C}{T} dt
$$

I take it by 'dt' you mean 'dT'?
I'm doing my best to get all of the thermodynamics we covered in Chemistry last semester clear in my head before we start it in Physics this semester. Would you mind casting an eye over this? I do get the same (correct) answer when rounded to 1 sig. figure but I'd like to be sure I'm doing this right:

(I've screwed up the formatting here a bit, sorry!)

$$
\Delta S = \int_{T_i}^{T_f} \frac{m C}{T} dT
$$
$$
= mC \int_{T_i}^{T_f} \frac{1}{T} dT
$$
= m⋅C⋅(ln Tf - ln Ti)
= (1 mol)(75 kJ⋅mol-1⋅K-1)(ln 323.15 K - ln 303.15 K)
= 4.79 kJ⋅K-1
 
  • #5
I haven't checked your arithmetic, but your approach is correct. Incidentally, ln(323.15)-ln(303.15)=ln(323.15/303.15). This will help you reduce roundoff error.
 
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  • #6
Chestermiller said:
I haven't checked your arithmetic, but your approach is correct. Incidentally, ln(323.15)-ln(303.15)=ln(323.15/303.15). This will help you reduce roundoff error.

Thanks Chester, I realized that just after posting - I think I have some revision to do re: logs, too...
 
  • #7
Ryaners said:
I take it by 'dt' you mean 'dT'?
Yes, that was a typo.
 
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Related to Calculating entropy change associated with change in temperature

1. What is entropy change?

Entropy change, also known as change in entropy, is a measure of the disorder or randomness of a system. It is a thermodynamic quantity that describes the amount of energy that is no longer available for work in a system.

2. How is entropy change calculated?

Entropy change is calculated by taking the integral of the change in heat divided by the temperature at constant pressure. This can be represented by the formula ΔS = ∫dq/T.

3. How does temperature affect entropy change?

Temperature plays a major role in determining the entropy change in a system. As temperature increases, the amount of disorder or randomness also increases, leading to a higher entropy change.

4. What is the relationship between temperature and entropy change?

The relationship between temperature and entropy change is directly proportional. This means that as temperature increases, the entropy change also increases.

5. Can entropy change be negative?

Yes, entropy change can be negative. This occurs when the disorder or randomness in a system decreases, leading to a decrease in entropy change. This is typically seen in processes that are more ordered, such as when a solid turns into a liquid or a liquid turns into a gas.

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