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. 1. The problem statement, all variables and given/known data 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. 2. Relevant equations q = m⋅C⋅ΔT ΔS = qrev / T (for a reversible process) 3. 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?