MT1-cheatsheet

# 025 wm c while there is still ice left nd the rate at

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Unformatted text preview: ) For one gas, say nitrogen, the volume increases by a factor of 3 and for the other the volume increases by a factor of 1.5. Thus we have SN = nR ln(V2N/V1) = nR ln(3) = nR ln 3. SA = nR ln(V2A/V1) = nR ln(1.5) = nR ln 1.5 . The total change of the system is Ssys = SN + SA = nR ln 3 + nR ln 1.5 = (1 mol)(8.315 J/mol K)(ln 3 + ln 1.5) =12.5 J/K. 20.58 (a) We find the final temperature from heat lost = heat gained; mwatercwater Twater = mAlcAl TAl ; (0.210 kg)(4186 J/kg C)(50C T) = (0.120 kg)(900 J/kg C)(T 15C), which gives T = 46.2C. (b) The heating and cooling do not occur at constant temperature. We add (integrate) the differential changes in entropy: Swater = (mwatercwater dT)/T = mwatercwater ln(T/Twater) = (0.210 kg)(4186 J/kg C) ln(319.4 K/323.2 K) = 10.4 J/K; SAl = (mAlcAl dT)/T = mAlcAl ln(T/TAl) = (0.120 kg)(900 J/kg C) ln(319.4 K/288.2 K) = + 11.1 J/K. The total entropy change is S = Swater + SAl = 10.4 J/K + 11.1 J/K = + 0.70 J/K. 20.66 (a) The actual efficiency of the engine is eactual = W/QH = (600 J)/(1600 J) = 0.375 = 37.5%. The ideal efficiency is eideal = 1 (TL/TH) = 1 [(400 K)/(850 K)] = 0.529 = 52.9%. Thus the engine is running at eactual /eideal = (0.375 )/(0.529 ) = 0.708 = 70.8% of ideal. (b) We find the heat exhausted in one cycle from QL = QH W = 1600 J 600 J = 1000 J. In one cycle of the engine there is no entropy change for the engine. The input heat is taken from the universe at TH and the exhaust heat is added to the universe at TL. The total entropy change is Stotal = ( QH/TH) + (QL /TL) = [ (1600 J)/(850 K)] + [(1000 J)/(400 K)] = + 0.618 J/K. (c) For a Carnot engine we have W = eQH = 0.529QH ; QL = QH W = (1 e)QH = (1 0.529 )QH = 0.471 QH . The total entropy change is Stotal = ( QH/TH) + (QL /TL) = [ QH/(850 K)] + [+ 0.471 QH/(400 K)] = 0, (d) For a Carnot engine we have WCarnot = eQH = 0.529QH = 0.529(1600 J) = 846 J. For the real engine Wreal = 600 J, so the difference is WCarnot Wreal = 846 J 600 J = 246 J. For TL S we get as expected for an id...
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## This note was uploaded on 09/09/2008 for the course PHYSICS 7B taught by Professor Packard during the Fall '08 term at University of California, Berkeley.

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