Question

some water is placed in a coffee-cup calorimeter. when 1.0g of an ionic solid is added, the temperature of the solution increases from 21.5C to 24.2C as the solid dissolves. For the dissolving process, what are the signs for delta Ssys,delta Ssurr, and delta Suniv? i dont understand it

Answer 1

Start with the easy one. Since the process was spontaneous, dS_u > 0. The Second Law tells you that. Now, your system released heat into the surroundings, which got hotter. Nothing *else* changed about the surroundings, we assume, so in this case you can use the Clausius relation to calculate the change of the entropy: dS_surr = dQ/T, where dQ is the heat that flowed in. So dS_surr > 0, also.

Finally, you're left with dS, the change in entropy of the salt itself. You know dS_u = dS + dS_surr, so that, at least, tells you that dS >= -dS_surr, so that dS_u >= 0. But that isn't enough to tell you the sign, since dS could either be positive or slightly negative.

I see no other way to proceed than to use your intuition that when an ordered solid, like a salt crystal, turns into a dispersed "cloud" of randomly placed and moving individual ions, the entropy will surely go up. So this tells you dS > 0, also.

Answer 2

I should add that in fact the entropy change of the solvent is very likely not just a result of the heat added to it by the dissolving salt. Other changes occur, which will modify the entropy change from what the Clausius relation predicts. The problem is you have a bit of a messy system/surroundings definition. I'm assuming the system is your salt, and the surroundings are the water in the cup. Unfortunately, that means the system is open to the surroundings, and more than just heat can flow between them.