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# chap 18 - HEAT WORK AND THE FIRST LAW OF THERMODYNAMICS 18...

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18.1 HEAT, WORK, AND THE FIRST LAW OF THERMODYNAMICS E XERCISES Section 18.1 The First Law of Thermodynamics 15. I NTERPRET We identify the system as the water in the insulated container. The problem is about work done to raise the temperature of a system. The first law of thermodynamics is involved. D EVELOP Since the container is perfectly insulated thermally, no heat enters or leaves the water in it. Thus, Q = 0 and the first law of thermodynamics in Equation 18.1 gives U Q W W = = − . The change in the internal energy of the water is determined from its temperature rise, U mc T = (see comments in Section 16.1 on internal energy). E VALUATE The work done on the water is W U mc T = − = − = − = − kg)(4.184 kJ/kg K)(7 K) (1 29 3 . kJ A SSESS The negative sign signifies that work was done on the water. 16. (a) The change in the internal energy of the water is U mc T = , and the work done by it (i.e., the negative of the work done on it) is given. Therefore, Equation 18.1 gives Q U W = + = + ( . 0 5 kg)(4.184 kJ/kg K)(3 K) ( . . . = = − 9 0 9 0 2 72 kJ) 6.28 kJ kJ kJ. (The negative sign signifies that the water lost heat to its surroundings.) (b) If the water had been in perfect thermal isolation, no heat would have been transferred, Q = 0 and W U = − = − 6 28 . kJ instead of 9 0 . kJ. 17. I NTERPRET We identify the system as the gas that undergoes expansion. The problem is about the change of internal energy of a system and involves the first law of thermodynamics. D EVELOP The heat added to the gas is Q Pt = = = (40 25 1000 W)( s) J. In addition, the amount of work it does on its surrounding is W = 750 J. The change in internal energy can be found by using the first law of thermodynamics given in Equation 18.1. E VALUATE Using Equation 18.1 we find U Q W = = = 1000 750 250 J J J A SSESS Since U > 0, we conclude that the internal energy has increased. 18. From Equation 18.2, dQ dt dU dt dW dt / / / = + = + = 45 W 165 W 210 W. 19. I NTERPRET This problem is about heat and mechanical energy, which are related by the first law of thermodynamics. The system is the automobile engine. D EVELOP Since we are dealing with rates, we make use of Equation 18.2: dU dt dQ dt dW dt = If we assume that the engine system operates in a cycle, then dU dt / . = 0 The engine's mechanical power output can then be calculated once the heat output is known. 18

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18.2 Chapter 18 E VALUATE The above conditions yield ( / ) dQ dt out kW = 68 and ( ) 0.17( ) in dW dt dQ dt / / = . Equation 18.2 then gives dW dt dQ dt dQ dt dQ dt d in out = = = 1 0 17 . W dt dQ dt out or dW dt dQ dt = = = ( / ) ( . ( . out ) kW ) 0 17 1 68 0 17 1 1 1 1 3 9 . kW A SSESS We find the mechanical power output dW dt / to be proportional to the heat output, ( / ) . dQ dt out In addition, dW dt / also increases with the percentage of the total energy released in burning gasoline that ends up as mechanical work.
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chap 18 - HEAT WORK AND THE FIRST LAW OF THERMODYNAMICS 18...

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