(10 blanks) Derive the thermal efficiency of a heat engine from the 1st law of thermodynamics. Since the process is cyclic, then AU = 0 so: Since: Q = QH+QC = |QH|-|QC| %3D %3D
(10 blanks) Derive the thermal efficiency of a heat engine from the 1st law of thermodynamics. Since the process is cyclic, then AU = 0 so: Since: Q = QH+QC = |QH|-|QC| %3D %3D
Physics for Scientists and Engineers: Foundations and Connections
1st Edition
ISBN:9781133939146
Author:Katz, Debora M.
Publisher:Katz, Debora M.
Chapter21: Heat And The First Law Of Thermodynamics
Section: Chapter Questions
Problem 11PQ
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![(10 blanks) Derive the thermal efficiency of a heat engine from the 1st law of thermodynamics.
Since the process is cyclic, then AU = 0
so:
=W
Since:
Q = QH+QC = |QH|-|QC|
Then
W =
But the efficiency
e=
/QH
Thus
+(
) =
- |
e =](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F6cc398c7-b0de-4d80-bcae-2b5a0da6e0a9%2F4d978144-f2c7-4d83-9f6a-1ebead72c30b%2Fmu1lezr_processed.png&w=3840&q=75)
Transcribed Image Text:(10 blanks) Derive the thermal efficiency of a heat engine from the 1st law of thermodynamics.
Since the process is cyclic, then AU = 0
so:
=W
Since:
Q = QH+QC = |QH|-|QC|
Then
W =
But the efficiency
e=
/QH
Thus
+(
) =
- |
e =
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