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Chap6_soln

# Chap6_soln - Problem set#6 4 9 10 32 4 Ok so the conversion...

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Problem set #6 4, 9, 10, 32 4. Ok, so the conversion of liquid water to ice can be summed up, fairly uninformatively, as H 2 O (l) H 2 O (s) But, I know that I have to heat up ice to get water, or, alternatively, cool down water to get ice. I can include this in my reaction informally, as H 2 O (l) H 2 O (s) + heat I put the heat on the right side because it can be treated just like another product: I have to pull heat out of liquid water to get ice, so ice and heat are both products, and both on the right side. Since the heat is on the right side, heat is being released, and so the reaction is exothermic, and I thus have energy going from my system (the ice/water) to the surroundings (a refrigerator perhaps… what does a fridge do? It sucks the heat out of things…). To think about kinetic and potential energy, we have to think about the molecules themselves: kinetic energy is still determined completely by temperature, so if had a water and ice at the same temperature ( 0 o C, for instance), then both water and ice will have the same kinetic energy. If the ice is colder, as it often is, then the frozen water molecules will have less kinetic energy. Now, lets go back to both samples being at 0.

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