FNT16 - FNT#17 Entropy Damien Martin Question 1 Calculate...

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FNT #17: Entropy Damien Martin November 27, 2006 Question 1 Calculate the change in entropy for the following situations: a) Heating 1.0 kg of solid ice from -100 C to 0 C. b) Melting 1.0 kg of solid ice at 0 C to liquid water at 0 C. c) Heating 1.0 kg of liquid water from 0 C to 100 C d) Vaporising 1.0 kg of water at 100 C to steam at 100 C. e) Rank these changes in entropy from lowest to highest. Which phase transition has the most entropy associated with it? a) Heating 1.0 kg of solid ice from -100 C to 0 C To be able to answer this question, let us look at the table on page 16 of our lab manual for the specific heat of ice. Taking into account the typographical error at the top of the column, we see that C p = 2 . 05 k J / (kg K). Because we only have 1 kg of ice, we know that the heat capacity is 2.05 kJ/K. Thus we have d Q = C d T, C = 2 . 05 kJ/K Recall that for small amounts of heat at constant temperature we have the relationship d S = d Q T which in this case is given by d S = d Q T = C d T T Δ S = T f T i C d T T 1
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If the heat capacity stays constant over the temperature range T i T T f then we can do this integral exactly: Δ S = C ln T f T i Now it is simply a case of putting in numbers. The only thing that we have to remember is that the temperatures must be given in Kelvin! Substituting, we find Δ S = (2 . 05 kJ/K) ln 273 173 = 0 . 935 kJ/K Where did all the signs go? After all, we did change the temperature of the ice, so heat had to be added in across temperature differences. Well, the idea is that we can do this very slowly: break the process up into millions of little pieces and slowly add heat to get a tiny change in temperature. In fact, this is exactly what the integral does! Of course, the ice does not really heat in this very slow idealised way in real life. But that is irrelevant! The point is that Δ S is a state function, so it depends only on the end points. c) Heating 1.0 kg of liquid water from 0 C to 100 C This is almost exactly the same as above: Δ S = C ln T f T i = (4 . 18 kJ/K) ln 373 273 = 1 . 305 kJ/K b) & d) The entropy changes at a phase change is even simpler to describe. The temperature is constant, and so we can use Q = | Δ H Δ m | if the process occurs at constant pressure . Recall that this is why we can use Δ H instead of heat. The change in entropy is given by Δ S = Q T = | Δ H Δ m | T 2
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Putting in numbers gives: Δ S ice water = 333 . 5 kJ 273 K = 1 . 22 kJ/K Δ S water vapor = 2257 kJ 373 K = 6 . 05 kJ/K Above I emphasised that this was valid at constant pressure. You may ask yourself – does it matter? After all, Δ S is a state function so it should not matter if the process is at constant pressure or constant volume – Δ S depends only on the endpoints. The answer is that it does matter. The statement that Δ S depends only on the initial and final states is completely true. If I picked two states and joined them by different processes, Δ S would be exactly the same. The problem here is to recall what “state” means in thermodynamics. For our ideal gases it means “having a particular N , V , P and T ”. Well, from the
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