SME
University Physics I: Mechanics · Problem type
The best a heat engine can do
Use this for “An engine runs between 500 K and 300 K and takes in 1000 J. At most how much work can it do?”
Diagram
QC = QH − W = 1,000 − 400
QC = 600 J
Efficiency = W/QH = 400/1,000 = 40%
Carnot limit = 1 − TC/TH = 1 − 300/500 = 40%
Some heat always goes to the cold reservoir: no engine turns all its heat into work.
This is an example. Type your own values to start a new problem.
e
%
=
TH (K)
−
TC (K)
TH (K)
QH
Heat taken in
example · J
W
Most work out
calculated · J
QC
Least heat thrown out
calculated · J
Formulas
e
=
100
×
TH − TC
TH
40
=
100
×
500 − 300
500
W
=
e
100
×
QH
400
=
40
100
×
1,000
QC
=
QH
−
W
600 = 1,000 − 400
Assumptions
•
A reversible (Carnot) engine; any real engine between the same two temperatures does less work: the second law.
•
Temperatures are in kelvins, the hot reservoir at TH and the cold one at TC.
•
Over one cycle the engine returns to its start (ΔU = 0), so the first law gives QH = W + QC.
Step-by-step
WE KNOW
TH = 500 K, TC = 300 K, QH = 1,000 J
FIND
Carnot efficiency (e), most work out (W), least heat thrown out (QC)
Step 1 · Find Carnot efficiency (e)
Use e = 100 × (TH − TC) ÷ TH
A
reversible
engine
throws
out
heat
in
the
ratio
QC
QH
=
TC
TH
,
so
the
share
left
as
work
is
1
−
TC
TH
.
e
=
100
×
500 − 300
500
e
=
100
×
200
500
e
=
20,000
500
e
=
40%
Step 2 · Find most work out (W)
Use W = e ÷ 100 × QH
The efficiency is the share of the heat in that comes out as work.
W
=
40
100
×
1,000
W
=
0.4
×
1,000
W
=
400
J
Step 3 · Find least heat thrown out (QC)
Use QC = QH − W
The engine ends each cycle as it began, so the heat not turned into work goes to the cold reservoir.
QC = 1,000 − 400
QC = 600 J
Check
Put the numbers back in. Both sides should match.
40
=
100
×
500 − 300
500
✓
400
=
40
100
×
1,000
✓
600 = 1,000 − 400 ✓
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First and second laws of thermodynamics
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The entropy gained heating water
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