CH 301 CH 9 answers - gs. surroundin on the system by the...

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Unformatted text preview: gs. surroundin on the system by the done is Work 0. w ie J 2280- w J/L.atm 101.325 30L atm 0.75- w 0. V P- w and 30L V thus V V 1 2 u u ! ' ' ! Chapter 9 answers to examples (some worked in class, some not) Example: If 1200 joules of heat are added to a system and the system does 800 joules of work on the surroundings, what is the energy change for the system, ' E " ' U = 1200J + (-800J) = + 400J What is the energy change of the surroundings? (-1200J +800J) = - 400J Example: Ideal gas is allowed to expand from a volume of 10L to a volume of 40L against a constant opposing pressure of 0.75atm. Find the work done. Example of Specific Heat: Amount of heat needed to raise the temperature of 150g of water from 15C to 35C. Specific heat of water is 4.184J/gC. Answer in Joules. 150g x 4.184J x 20C = 12,540J gC +HDWLQJ &RROLQJ FXUYH H[DPSOHV ' H fus of benzene is 10.59 kJ/mol. I have 47g of liquid benzene at its melting point, 278.6K. If I remove 2000J of heat energy from this sample, how much liquid benzene is left? ' H freezing = - ' H fusion moles benzene 47g X 1mol/78g = 0.60mol q = - 2kJ = - 10.59kJ/mol * ?mol = 0.188 moles of benzene freeze so, 0.6 - 0.188 moles = 0.411 = 32 g left as liquid. What will be the final temperature of 20L of water at room temperature (25.000C) if 200g of ice at 1 C is dropped into it? What happens? BREAK IT INTO STEPS! 1. 200g ice at 1 C warms to 0C: Heat energy flows from water to ice: q (into ice) = 200g * (2.03J/g C) * (1 C) = 406J absorbed from warm water. 2. Warm water cools slightly as result: q (out of warm water) = -q (into ice) = - 406J If - 406J = 20L * 1000g/1L * 4.184J/g C * ' T this gives ' T = 0.00485 C Water is now at (25 - 0.00485) C = 24.995 C 3. 200g ice at 0 C melts: Heat energy flows from water to ice: q (into ice) = 200g/18(g/mol) * 6.01kJ/mol= 66777.8J absorbed from warm water. 4. Warm water cools further as a result: q (out of warm water) = -q (into ice) = - 66777.8J If - 66777.8J = 20L * 1000g/1L * 4.184J/g C * ' T this gives ' T = 0.7980 C Water is now at (24.995 - 0.7980 ) C = 24.197 C 5. Now have 20g water at 0 C next to warmer water at 24.197 C. Heat flows from warmer water to cooler water until they are BOTH at the same final temp: q (into cold water) = - (q (out of warm water)) WATCH ALGEBRA!!!!! 200g * 4.184J/g C * (T f- 0) = - [20L * 1000g/1L * 4.184J/g C * (T f- 24.197) ] divide both sides by (200 * 4.184) to get: ( T f - 0) = -[100 (T f- 24.197)] T f = - 100 T f + 2419.7 101T f = 2419.7 T f = 23.957 C ie dropped it about a degree. Compare the energy given out to your arm if its exposed to: A. 5g of boiling water. B. 5g of steam Assume water ends up at body temperature, about 38 C ' H fus water = 6.01kJ/mol ' H vap water = 40.7kJ/mol C s water = 4.184 J/g C A: water at 100 C, cools to 38 C....
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This note was uploaded on 02/12/2012 for the course CH 301 taught by Professor Fakhreddine/lyon during the Summer '07 term at University of Texas at Austin.

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CH 301 CH 9 answers - gs. surroundin on the system by the...

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