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Unformatted text preview: Lecture 15, p 1 Lecture 15 Heat Engines Review & Examples Hot reservoir at T h Cold reservoir at T c Q h Q c W Heat pump Heat leak The cycle goes around like this. We don’t describe the mechanical parts that move the gas cylinder back and forth between the reservoirs. Lecture 15, p 2 To analyze heat engines, we need to be able to calculate ∆ U, ∆ T, W, Q, etc . for the processes that they use. How much heat is absorbed by 3 moles of helium when it expands from V i = 10 liters to V f = 20 liters and the temperature is kept at a constant 350 K ? What are the initial and final pressures? Example Process Lecture 15, p 3 To analyze heat engines, we need to be able to calculate ∆ U, ∆ T, W, Q, etc . for the processes that they use. How much heat is absorbed by 3 moles of helium when it expands from V i = 10 liters to V f = 20 liters and the temperature is kept at a constant 350 K ? What are the initial and final pressures? Solution Q = W by The 1 st law. For an ideal gas, ∆ T = 0 → ∆ U = 0. Positive Q means heat flows into the gas. W by = nRT ln(V f /V i ) = 6048 J An expanding gas does work. p i = nRT/V i = 8.72 × 10 5 Pa Use the ideal gas law, pV = nRT p f = p i /2 = 4.36 × 10 5 Pa Where is the heat coming from? In order to keep the gas at a constant temperature, it must be put in contact with a large object ( a “heat reservoir” ) having that temperature. The reservoir supplies heat to the gas (or absorbs heat, if necessary) in order to keep the gas temperature constant. Very often, we will not show the reservoir in the diagram. However, whenever we talk about a system being kept at a specific temperature, a reservoir is implied. Lecture 15, p 4 Suppose a mole of a diatomic gas, such as O 2 , is compressed adiabatically so the final volume is half the initial volume . The starting state is V i = 1 liter , T i = 300 K . What are the final temperature and pressure? Example Process (2) Lecture 15, p 5 6 1 7/ 2 1.4 5/ 2 6.57 10 Pa 395 K 395 K i i f f i f i f i i i f f f f i i f f i f i f pV pV V p p V nRT V V V x pV T nR T V T V V T T V γ γ γ γ α α α γ = = = = = = = = = = = Suppose a mole of a diatomic gas, such as O 2 , is compressed adiabatically so the final volume is half the initial volume . The starting state is V i = 1 liter , T i = 300 K . What are the final temperature and pressure? Equation relating p and V for adiabatic process in αideal gas γ is the ratio of C p /C V for our diatomic gas ( =( α +1)/ α ). Solve for p f . We need to express it in terms of things we know. Use the ideal gas law to calculate the final temperature. Alternative: Use the equation relating T and V for an adiabatic process to get the final temperature. α = 5/2 Solution Lecture 15, p 6 Two blocks, each with heat capacity* C = 1 J/K are initially at different temperatures, T 1 = 250 K , T 2 = 350 K ....
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This note was uploaded on 02/21/2011 for the course PHYS 213 taught by Professor Staff during the Spring '08 term at University of Illinois, Urbana Champaign.
 Spring '08
 Staff
 Heat

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