thermoanswers

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

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
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
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

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....
View Full Document

This note was uploaded on 07/16/2008 for the course MAE 101 taught by Professor Lee during the Spring '08 term at ASU.

Ask a homework question - tutors are online