Ch20Word - Chapter 20 CHAPTER 20 Second Law of...

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Chapter 20 Page 1 CHAPTER 20 – Second Law of Thermodynamics; Heat Engines 1. For the heat input, we have Q H = Q L + W = 8500 J + 2700 J = 11,200 J. We find the efficiency from e = W / Q H = (2700 J)/(11,200 J) = 0.24 = 24% . 2. We find the efficiency from e = W / Q H = (8200 J)/(18.0 kcal)(4186 J/kcal) = 0.109 = 10.9% . 3. We find the rate of heat input from the efficiency: e = W / Q H = ( W / t )/( Q H / t ); 0.38 = (500 MW)/( Q H / t ), which gives Q H / t = 1316 MW. We find the rate of heat discharge from Q H / t = ( Q L / t ) + ( W / t ); 1316 MW = ( Q L / t ) + 500 MW, which gives Q L / t = 816 MW . 4. ( a ) We find the rate at which work is done from P = W / t = (180 J/cycle/cyl)(4 cyl)(25 cycles/s) = 1.8 × 10 4 J/s . ( b ) We find the heat input rate from e = W / Q H = ( W / t )/( Q H / t ); 0.25 = (1.8 × 10 4 J/s)/( Q H / t ), which gives Q H / t = 7.2 × 10 4 J/s . ( c ) We find the time to use one gallon from t = E /( Q H / t ) = (130 × 10 6 J/gal)(1 gal)/(7.2 × 10 4 J/s) = 1.81 × 10 3 s = 30 min . 5. We use the units to help us find the rate of heat input to the engine from the burning of the gasoline: Q H / t = [(3.0 × 10 4 kcal/gal)/(38 km/gal)](90 km/h)(4186 J/kcal) = 2.97 × 10 8 J/h. The horsepower is the work done by the engine. We find the efficiency from e = W / Q H = (20 hp)(746 W/hp)(3600 s/h)/(2.97 × 10 8 J/h) = 0.18 = 18% . 6. ( a ) Work is positive for an expansion. Because the work done is represented by the area under the PV curve, W bc > W ac . The net work is the sum of the works done in each leg. For a positive net work, W bc > 0, so we have an expansion from b to c, and the path must be clockwise. ( b ) We find the ratio of volumes from the ideal gas equation: ( P c / P a )/( V c / V a ) = T c / T a ; (1)( V c / V a ) = T c / T a , so V c / V a = V c / V b = T c / T a = T b / T a . The work done during the isothermal expansion is W bc = nRT b ln( V c / V b ) = nRT b ln( T b / T a ) = (1.0 mol)(8.315 J/mol · K)(423 K) ln(423 K/273 K) = 1.54 × 10 3 J. The work done during the constant pressure compression is W ca = P a ( V a V c ) = P a V a [1 – ( V c / V a )] = nRT a [1 – ( V c / V a )] = (1.0 mol)(8.315 J/mol · K)(273 K)[1 – (423 K/273 K)] = – 1.25 × 10 3 J. For the isothermal expansion ? U bc = 0, so Q bc = W bc = 1.54 × 10 3 J. For the constant volume compression W ab = 0, so Q ab = ? U ab = nC V ( T b T a ) = (1.0 mol) * (8.315 J/mol · K)(423 K – 273 K) = 1.87 × 10 3 J. The efficiency is e = W net / Q added = ( W bc + W ca )/( Q bc + Q ab ) = ( 1 . 5 4 × 10 3 J – 1.25 × 10 3 J)/(1.54 × 10 3 J + 1.87 × 10 3 J) = 0.085 = 8.5% . P V b a 0 c T b = T c
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Chapter 20 Page 2 7. We find the heat input rate from e = W / Q H ; 0.38 = (810 MW)/( Q H / t ), which gives Q H / t = 2132 MW. We find the discharge heat flow from Q L / t = ( Q H / t ) – ( W / t ) = 2132 MW – 810 MW = 1322 MW. If this heat flow warms the air, we have Q L / t = ( n / t ) c P ? T ; (1322 × 10 6 W)(3600 s/h)(24 h/day) = ( n / t )(7.0 cal/mol · C°)(7.5 C°)(4.186 J/cal), which gives n / t = 5.20 × 10 11 mol/day. To find the volume rate, we use the ideal gas law: P ( V / t ) = ( n / t ) RT ; (1.013 × 10 5 Pa)( V / t ) = (5.20 × 10 11 mol/day)(8.315 J/mol · K)(293 K), which gives V / t = 1.25 × 10 10 m 3 /day = 13 km 3 /day . Depending on the dispersal by the winds, the local climate could be heated significantly. We find the area from A = ( V / t ) t / h = (12.5 km 3 /day)(1 day)/(0.200 km) = 63 km 2 . 8. ( a ) We find the heat flow and work done for each leg.
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This note was uploaded on 02/05/2010 for the course PHYSICS 37,38,39 taught by Professor Fielding during the Spring '09 term at Los Angeles Valley College.

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Ch20Word - Chapter 20 CHAPTER 20 Second Law of...

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