# thermoanswers - “roach motel” for vapors perhaps –...

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Answers to sample questions 1.1., 1.2 addressed in class Wed 1.3. A) Standard calorimetry exercise: q = m x C x ) T – the same amount of heat lost by the metal is used to heat the water Water: m = 150 g, C = 4.18 J/g.K, ) T = 32.3 - 25 = 7.3 K Metal m = 42 g, ) T = 165 - 32.3 = 132.7 K C(metal) = q(water)/(42 g x 132.7 K) B) q to cool 500 g of water by 25 K is 500 g x 4.18 J/g.K x 25 K ~ 50,000 J Heat of fusion of ice is 333 J/g; (Note that the melting ice removes heat from the water) No. Of g of ice needed to remove 50,000 J of heat is 50,000 J/333 J/g 2.1 a Invention violates 2 Law of Thermodynamics. Reconcentrating a diffuse vapor violates nd the idea that gases spontaneously become more diffuse (entropy increases). (On recollection, this question has some unseen subtleties: for example, if we make a device – a
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Unformatted text preview: “roach motel” for vapors perhaps – that traps any vapor molecule that happens randomly to diffuse in, that would remove the odor without violating the 2 law. We would be using an nd enthalpy term – some enthalpy of binding of an odor molecule to the walls of the trap – to overcome the tendency of the molecules to become increasinngly diffuse. This would be an example of the ) H term in ) G overcoming the T ) S term.) b) Worked in class. ) S = q/T 2.2 a) Entropy change is S(reactants) - S(products) = 2(240) + 2(70) - 121 - 3(205) b) If the hydrazine was a gas, its entropy of formation would be significantly higher (several hundred J/K.mol – compare O2, NO2) so the net increase in entropy would be less....
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## This note was uploaded on 07/16/2008 for the course MAE 101 taught by Professor Lee during the Spring '08 term at ASU.

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