Calculating entropy change associated with change in temperature

[Ryaners]

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 = q_rev / 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?

 

[DrClaude]

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:
$$
\Delta S = \int_{T_i}^{T_f} \frac{m C}{T} dT
$$

 

[Ryaners]

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 :)

 

[Ryaners]

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 T_f - ln T_i)
= (1 mol)(75 kJ⋅mol-1⋅K-1)(ln 323.15 K - ln 303.15 K)
= 4.79 kJ⋅K^-1

 

[Chestermiller]

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.

 

[Ryaners]

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...

 

[DrClaude]

I take it by 'dt' you mean 'dT'?

Yes, that was a typo.