Question

Problem 1 (10 pts) (a) A heat engine operates between a reservoir at 350°C, from which it absorbs 200 kJ of heat per cycle, and a reservoir at 0°C, to which it rejects 115 kJ of heat per cycle. Determine if this engine represents a reversible, irreversible, or impossible cycle? (4 pts) (b) An insulated container of capacity 2 m³ contains 10 kg of air. Work is done on the air till its pressure increases from 500 KPa to 1 MPa. Determine the change in entropy. Assume that specific heats evaluated at 300 K. (4 pts) (c) Find the maximum possible thermal efficiency for a heat engine operating between two reservoirs at 750°C and 27°C. (2 pts)

          Problem 1 (10 pts)
(a) A heat engine operates between a reservoir at 350°C, from which it absorbs 200 kJ of heat per cycle, and a reservoir at 0°C, to
which it rejects 115 kJ of heat per cycle. Determine if this engine represents a reversible, irreversible, or impossible cycle? (4 pts)
(b) An insulated container of capacity 2 m³ contains 10 kg of air. Work is done on the air till its pressure increases from 500 KPa
to 1 MPa. Determine the change in entropy. Assume that specific heats evaluated at 300 K. (4 pts)
(c) Find the maximum possible thermal efficiency for a heat engine operating between two reservoirs at 750°C and 27°C. (2 pts)

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Problem 1 (10 pts)
(a) A heat engine operates between a reservoir at 350°C, from which it absorbs 200 kJ of heat per cycle, and a reservoir at 0°C, to
which it rejects 115 kJ of heat per cycle. Determine if this engine represents a reversible, irreversible, or impossible cycle? (4 pts)
(b) An insulated container of capacity 2 m³ contains 10 kg of air. Work is done on the air till its pressure increases from 500 KPa
to 1 MPa. Determine the change in entropy. Assume that specific heats evaluated at 300 K. (4 pts)
(c) Find the maximum possible thermal efficiency for a heat engine operating between two reservoirs at 750°C and 27°C. (2 pts)

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University Physics with Modern Physics

Hugh D. Young 14th Edition

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Solved by Expert Devya P

02/02/2022

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Step 1: The efficiency of a heat engine is defined as the ratio of the work done by the engine to the heat absorbed from the hot reservoir. Show more…

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Problem 1 (10 pts) (a) A heat engine operates between a reservoir at 350°C, from which it absorbs 200 kJ of heat per cycle, and a reservoir at 0°C, to which it rejects 115 kJ of heat per cycle. Determine if this engine represents a reversible, irreversible, or impossible cycle? (4 pts) (b) An insulated container of capacity 2 m³ contains 10 kg of air. Work is done on the air till its pressure increases from 500 KPа to 1 MPa. Determine the change in entropy. Assume that specific heats evaluated at 300 K. (4 pts) (c) Find the maximum possible thermal efficiency for a heat engine operating between two reservoirs at 750°C and 27°C. (2 pts)

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Transcript

00:01 So, the given part of the question is q that enters inside the system is given 3 ,500 jule, q that leaves out or released from the system in the form of heat is given 840 jule, temperature is given as 1 ,500 kelvin.

00:22 In the first part of the question, we have to find delta s, that is entropy change of the system.

00:27 And as we all know that for a reversible process, reversible process, dail q is equal to t, dales s.

00:43 So from here, after substituting values, we will find dale s is equal to dale q upon t.

00:53 Heat of the system is 3 ,500 upon temperature is 1500.

01:01 After calculating the values, we will find 7 by 3 joules.

01:05 Per kelvin.

01:07 This is the value of delta s.

01:10 And after further more simplification, we will find value of 2 .3333 and so on, jule per kelvin.

01:18 Taking an approximate value, dale s comes out to be 2 .33 joule per kelvin.

01:27 So this is the answer of our part a.

01:32 To calculate the part b, we have to calculate entropy change in an adiabetic process.

01:38 And as we all know that in an adiabetic process, no heat flows between the system and the surrounding.

01:45 So, d .l .q is zero.

01:49 Again coming out to the same formula is d .l.

01:52 Q is equal to t, dales, s.

01:55 And after substituting the values, we'll get dals is equal to dl q upon.

02:04 And as said, dl q is 0...

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