An irreversible engine operates between temperatures of 859 and 312 K. It absorb

Question

An irreversible engine operates between temperatures of 859 and 312 K. It absorbs 1288 J of heat from the hot reservoir and does 286 J of work. (a) What is the change ?Suniverse in the entropy of the universe associated with the operation of this engine? (b) If the engine were reversible, what would be the magnitude |W| of the work it would have done, assuming that it operated between the same temperatures and absorbed the same heat as the irreversible engine? (c) Using the results of parts (a) and (b), find the difference between the work produced by the reversible and irreversible engines.

Explanation / Answer

in this problem process is irreversible

so that engine will have less efficiency

first we need to find out efficiency

EFFICIENCY OF REVERSIBLE PROCESS

=1-(TL/TH)

=1-0.363

=0.636

=63.6 PERCENT

IN IRREVERSIBLE ENGINE NET WORK DONE EQUALS TO

QH=QL+W

1288=QL+286

QL=1002 JOULES

EFFICIENCY OF ENGINE IN IRREVERSIBLE PROCESS

=1-(QL/QH)

=1-0.779

=0.222

=22.2 PERCENT

ENTROPY

DELTA SL=-QL/TL

=-3.21J/K

DELTA SH   =QH/TH

=1288/859

=1.49J/K

TOTAL ENTROPY IS IN ZERO IN IRREVERSIBLE CASE

IN REVERSIBLE CASE

LET US CONSIDER ENGINE PERFORMS 286J WORK

FIND ENERGY ABSORBED FROM HOT RESERVIOR

QH=W/EFFICIENCY

=451.1J

USE RELATION QH=QL+W

QL  =165.1J

FIND ENTROPY

DELTA SL=-0.529

DELTA SH=0.525

SO IN THIS PROCESS ENTROPY ALWAYS EQUAL TO ZERO   

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