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

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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....
View Full Document

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.

Page1 / 18

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

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online