- #1

Ryaners

- 50

- 2

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?