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# Entropy and the Second Law of Thermodynamics.

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Description
Heat Engine and Refrigerators. A warmth motor is a gadget that helps a working substance through a cyclic procedure, amid which it. Retains warm vitality from a high-temperature source (H). The motor works and . Removes warm vitality to a low-temperature source (L). . first Law. Cyclic Process.
Transcripts
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﻿Entropy and the Second Law of Thermodynamics

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Heat Engine and Refrigerators A warmth motor is a gadget that brings a working substance through a cyclic procedure, amid which it Absorbs warm vitality from a high-temperature source (H) The motor does work and Expels warm vitality to a low-temperature source (L)

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Work done by the substance Heat added to the substance Thermal proficiency: first Law Cyclic Process

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A cooler is a warmth motor backward. A cooler is a gadget that brings a working substance through a cyclic procedure, amid which it Absorbs warm vitality from a low-temperature source by doing Work Expels warm vitality (the sum work done and the warmth retained) to a high-temperature source

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Heat consumed by the substance Work done on the substance Coefficient of execution: first Law Cyclic Process

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Reversible and Irreversible Processes Reversible Process : a procedure that at the conclusion, the framework and its encompassing come back to the correct beginning conditions. Every characteristic procedure are irreversible, best case scenario "practically" reversible .

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Irreversible Processes Heat streams from the high-temperature to the low-temperature objects. The ricocheting ball at last stops The wavering pendulum at long last stops

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Carnot Engine (an Ideal Engine) Carnot motor: a warmth motor working in a Carnot cycle . Carnot cycle: a cycle comprising of four REVERSIBLE (hence Ideal) forms : two adiabatic procedures B C; D An and two isothermal procedures. A B;C D Carnot motor is the most effective warmth motor conceivable.

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B  C D  An Efficiency of a Carnot Engine Isothermal: A  B C  D Adiabatic:

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Efficiency of an Ideal Carnot Engine Thermal effectiveness is then

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Performance of an Ideal Carnot Refrigerator Coefficient of execution

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T - T ( 235 - 115 ) K H C e = 23 . 6 % T ( 235 + 273 ) K H (a) HRW 38E (5 th ed.) . A perfect warmth motor works in a Carnot cycle somewhere around 235˚C and 115˚C. It assimilates 6.30x10 4 J for each cycle at the higher temperature. (a) What is the effectiveness of the motor? (b) How much work per cycle is this motor fit for performing? (a)

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c b V, p (a) Pressure V 0 , p 0 d a Volume HRW 45P (5 th ed.) . One mole of a perfect monatomic gas is taken through the cycle appeared. Expect that p =2 p 0 , V =2 V 0 , p 0 = 1.01x10 5 Pa, and V 0 = 0.0225 m 3 . Ascertain (a) the work done amid the cycle, (b) the warmth included amid stroke abc , and (c) the proficiency of the cycle. (d) What is the proficiency of a perfect motor working between the most elevated and least temperatures that happen in the cycle? How does this contrast with the productivity computed in (c)? (b)

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c (b) V, p Pressure V 0 , p 0 d a (c) Volume HRW 45P (5 th ed.) . One mole of a perfect monatomic gas is taken through the cycle appeared. Expect that p =2 p 0 , V =2 V 0 , p 0 = 1.01x10 5 Pa, and V 0 = 0.0225 m 3 . Compute (a) the work done amid the cycle, (b) the warmth included amid stroke abc , and (c) the productivity of the cycle. (d) What is the effectiveness of a perfect motor working between the most elevated and least temperatures that happen in the cycle? How does this contrast with the productivity computed in (c)?

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c b V, p (d) Pressure V 0 , p 0 d a Volume HRW 45P (5 th ed.) . One mole of a perfect monatomic gas is taken through the cycle appeared. Accept that p =2 p 0 , V =2 V 0 , p 0 = 1.01x10 5 Pa, and V 0 = 0.0225 m 3 . Figure (a) the work done amid the cycle, (b) the warmth included amid stroke abc , and (c) the effectiveness of the cycle. (d) What is the productivity of a perfect motor working between the most astounding and least temperatures that happen in the cycle? How does this contrast with the effectiveness ascertained in (c)? Bigger than 15%.

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In one hour HRW 60P (5 th ed.) . A perfect warmth pump is utilized to warm a building. The outside temperature is - 5.0˚C and the temperature inside the building is to be kept up at 22˚C. The coefficient of execution is 3.8 and the warmth pump conveys 7.54 MJ of warmth to the building every hour. At what rate must work be done to run the warmth pump? For a total cycle, ∆E int = 0 Rate = 1.57 x 10 6 J/3600 s = 440 W

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Entropy and the Second Law of Thermodynamics

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The Second Law of Thermodynamics It is difficult to build a warmth motor with 100% proficiency. It is difficult to build an icebox that does not require work.

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Absolute Temperature Scale Carnot motor — the most productive warmth motor: T C = 0  e = 1 Second Law  Temperature can\'t be equal or lower than T = 0 K

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The subscript "r" remains for reversible process Entropy is a state work like p , T , and E int . A state work depicts the thermodynamic condition of a framework, which is free of the history . The adjustment in entropy S for a little procedure:

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The combination is along a way that speaks to a reversible procedure For a limited procedure from an underlying state " i " to a last state " f " , the adjustment in entropy is Unit: J/K

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From the factual mechanical perspective , entropy is a measure of turmoil .

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No vitality trade with different frameworks The Second Law of Thermodynamics In a CLOSED framework: = : reversible process > : irreversible process

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Entropy Carnot Engine For the gas : the Carnot cycle is a reversible shut cycle — back to a similar state: dQ ò D = Ñ = r S 0 T

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Isothermal Carnot Engine For the hot supply: warmth is lost A  B For the frosty repository : warmth is picked up C  D

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Carnot cycle Carnot Engine B  C and D  An: Adiabatic

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HRW 5E (5 th ed.) . A perfect gas in contact with a steady temperature supply experiences a reversible isothermal development to twice its underlying volume. Demonstrate that the supply\'s adjustment in entropy is free of its temperature. Isothermal ∆E int = 0  Q = W Change of entropy of the perfect gas: Reversible process: ∆S add up to = 0 Change of entropy of the reservoir: ∆S\' = - ∆S = - nR ln2

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a. b. Monatomic 400 Temperature (K) 200 c. 5 10 15 20 Entropy (J/K) HRW 15P (5 th ed.) . A 2.0 mol test of a perfect monatomic gas experiences the reversible procedure appeared. (a) How much warmth is consumed by the gas? (b) What is the adjustment in the inner vitality of the gas? (c) How much work is finished by the gas?

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HRW 25P (5 th ed.) . A 10 g ice solid shape at - 10˚C is put in a lake whose temperature is 15˚C. Ascertain the adjustment in entropy of the shape lake framework as the ice 3D square comes to warm balance with the lake. The particular warmth of ice is 2220 J/kg·K. The ice warms from - 10˚C to 0˚C (1), then melts (2), then the water warms to the lake temperature of 15˚C (3). The entropy change of the ice: (1)

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(2) (3) HRW 25P (5 th ed.) . A 10 g ice solid shape at - 10˚C is set in a lake whose temperature is 15˚C. Figure the adjustment in entropy of the 3D shape lake framework as the ice block comes to warm harmony with the lake. The particular warmth of ice is 2220 J/kg·K.

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(1) (2) (3) HRW 25P (5 th ed.) . A 10 g ice solid shape at - 10˚C is set in a lake whose temperature is 15˚C. Compute the adjustment in entropy of the 3D shape lake framework as the ice block comes to warm harmony with the lake. The particular warmth of ice is 2220 J/kg·K. The entropy change of the lake: expect its temperature does not change: ∆S L = Q/T

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HRW 25P (5 th ed.) . A 10 g ice solid shape at - 10˚C is set in a lake whose temperature is 15˚C. Compute the adjustment in entropy of the 3D shape lake framework as the ice 3D square comes to warm balance with the lake. The particular warmth of ice is 2220 J/kg·K.

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