18_InstSolManual_PC

18_InstSolManual_PC - HEAT WORK AND THE FIRST LAW OF...

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Unformatted text preview: 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 = 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 (1 kg)(4.184 kJ/kg K)(7 K) 29.3 kJ W U mc T = -∆ = - ∆ = - ⋅ = - 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 (0.5 kg)(4.184 kJ/kg K)(3 K) Q U W = ∆ + = ⋅ + ( 9.0 kJ) 6.28 kJ 9.0 kJ 2.72 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 = and 6.28 kJ W U = -∆ = - 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 (40 W)(25 s) 1000 J. Q Pt = = = In addition, the amount of work it does on its surrounding is 750 J. W = 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 1000 J 750 J 250 J U Q W ∆ =- =- = A SSESS Since 0, U ∆ we conclude that the internal energy has increased. 18. From Equation 18.2, / / / 45 W 165 W 210 W. dQ dt dU dt dW dt = + = + = 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 dQ dW dt dt dt =- If we assume that the engine system operates in a cycle, then / 0. dU dt = The engine's mechanical power output can then be calculated once the heat output is known. E VALUATE The above conditions yield out ( / ) 68 kW dQ dt = and in ( / ) 0.17( / ) . dW dt dQ dt = Equation 18.2 then gives 1 0.17 in out out dW dQ dQ dQ dW dQ dt dt dt dt dt dt = =- =- 18.1 18 18.2 Chapter 18 or out 1 1 ( / ) 68 kW 13.9 kW (0.17) 1 (0.17) 1 dQ dt dW dt-- = = =-- A SSESS We find the mechanical power output / dW dt to be proportional to the heat output, out ( / ) . dQ dt In addition, / dW dt also increases with the percentage of the total energy released in burning gasoline that ends up as mechanical work.mechanical work....
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This note was uploaded on 09/12/2009 for the course PH PH101 taught by Professor Prof.kim during the Spring '09 term at Korea Advanced Institute of Science and Technology.

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18_InstSolManual_PC - HEAT WORK AND THE FIRST LAW OF...

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