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assign6sol

# assign6sol - 3 C K(100 kg J kg(390 50 3 ° = ⋅ ⋅ ° ⋅...

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Solutions for Assignment 6 1: a) %. 9 . 28 289 . 0 b) J. 2600 J 6400 J 9000 J 9000 J 2600 = = = - 2: a) MW. 970 MW 330 MW 1300 b) %. 25 25 . 0 MW 1300 MW 330 = - = = = e 3: a) ) ( water water ice ice C T c T c L m Q f D + D + = ( ) J. 10 90 . 8 K) K)(25.0 kg J 4190 ( K) K)(5.0 kg J 2100 ( kg J 10 334 kg) 80 . 1 ( 5 3 = + + = b) J. 10 37 . 3 5 40 . 2 J 10 08 . 8 | | 5 C = = = K Q W c) = = + = + = | | that (note J 10 1.14 J 10 8.08 J 10 37 . 3 | | | | H 6 5 5 C H Q Q W Q ).) 1 ( | | 1 C K Q + 4: The claimed efficiency of the engine is . % 58 J 10 60 . 2 J 10 51 . 1 8 8 = While the most efficient engine that can operate between those temperatures has efficiency %. 38 1 K 400 K 250 Carnot = - = e The proposed engine would violate the second law of thermodynamics, and is not likely to find a market among the prudent. 5: a) The final temperature, found using the methods of Chapter 17, is C, 94 . 28 K) kg J kg)(4190 (0.800 K) kg J kg)(390 50
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Unformatted text preview: . 3 ( ) C K)(100 kg J kg)(390 50 . 3 ( ° = ⋅ + ⋅ ° ⋅ = T or C 9 . 28 ° to three figures. b) Using the result of Example 20.10, the total change in entropy is (making the conversion to Kelvin temperature) K. J 2 . 49 K 273.15 K 302.09 ln K) kg J kg)(4190 800 . ( K 373.15 K 302.09 ln K) kg J kg)(390 50 . 3 ( = ˜ ˜ ˆ ⋅ + ˜ ˆ ⋅ = D S (This result was obtained by keeping even more figures in the intermediate calculation. Rounding the Kelvin temperature to the nearest K 01 . gives the same result....
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