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

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18.1 HEAT, WORK, AND THE FIRST LAW OF THERMODYNAMICS EXERCISES Section 18.1 The First Law of Thermodynamics 15. INTERPRET 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. DEVELOP 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 UQW W =− = . The change in the internal energy of the water is determined from its temperature rise, ∆∆ Um cT = (see comments in Section 16.1 on internal energy). EVALUATE The work done on the water is WU m c T =− kg)(4.184 kJ/kg K)(7 K) (1 293 .k J ASSESS The negative sign signifies that work was done on the water. 16. (a) The change in the internal energy of the water is = , and the work done by it (i.e., the negative of the work done on it) is given. Therefore, Equation 18.1 gives QU W =+ = + (. 0 5 kg)(4.184 kJ/kg K)(3 K) . . −= 90 272 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 628 J instead of J . 17. INTERPRET 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. DEVELOP The heat added to the gas is QP t == = (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. EVALUATE Using Equation 18.1 we find UQW = 1000 750 250 JJJ ASSESS 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. INTERPRET This problem is about heat and mechanical energy, which are related by the first law of thermodynamics. The system is the automobile engine. DEVELOP 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 EVALUATE 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 == = 1 017 . W dt dQ dt out or dt dQ dt = = = −− ) (. out ) kW ) 1 68 1 1 11 39 .kW ASSESS 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. Section 18.2 Thermodynamic Processes 20. The work done by the gas equals the area under the straight diagonal path AB in Fig. 18.19. The area of this trapezoid is WP P V V P P V V P V =+ = 1 2 12 21 1 2 1 1 3 2 1 22 () ( ) ( ) 1 .
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This note was uploaded on 10/21/2010 for the course PHYSICS 2131441 taught by Professor Pheong during the Fall '10 term at University of California, Berkeley.

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

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