SME
Grade 11 Science · Problem type
A refrigerator
Use this for “A fridge removes 900 J from its 270 K inside using 300 J of work. How much heat reaches the 300 K kitchen, and what are its COP and the best possible COP?”
Diagram
Qₕ = Qc + W = 900 + 300 = 1,200 J
COP = Qc/W = 900/300 = 3
Carnot COP = Tc/(Tₕ − Tc) = 270/(300 − 270) = 9
Work moves heat from cold to hot, which never happens by itself.
This is an example. Type your own values to start a new problem.
Qc
Heat taken from inside
example · J
W
Work
example · J
Qₕ
Heat given to the room
calculated · J
COP
Coefficient of performance
calculated
Tₕ
Hot reservoir temperature
example · K
Tc
Cold reservoir temperature
example · K
COPC
Carnot COP
calculated
r
Share of the Carnot COP
calculated · %
Formulas
Qₕ
=
Qc
+
W
1,200 = 900 + 300
COP
=
Qc/
W
3
=
900
300
COPC = Tc/(Tₕ − Tc)
9 = 270/(300 − 270)
r
=
100
×
COP/COPC
33.3
=
100
×
3
9
Assumptions
•
Heat flows from hot to cold by itself; moving it the other way takes work.
•
The kitchen gets the heat from inside plus the work: Qₕ = Qc + W.
•
COP = Qc/W; the best possible (Carnot) COP is Tc/(Tₕ − Tc), in kelvins.
Step-by-step
WE KNOW
Qc = 900 J, W = 300 J, Tₕ = 300 K, Tc = 270 K
FIND
heat given to the room (Qₕ), coefficient of performance (COP), Carnot COP (COPC), share of the Carnot COP (r)
Step 1 · Find heat given to the room (Qₕ)
Use Qₕ = Qc + W
The room gets the heat from inside plus the work.
Qₕ = 900 + 300
Qₕ = 1,200 J
Step 2 · Find coefficient of performance (COP)
Use COP = Qc/W
Divide the heat moved by the work.
COP
=
900
300
COP = 3
Step 3 · Find Carnot COP (COPC)
Use COPC = Tc/(Tₕ − Tc)
The Carnot COP: the cold temperature over the difference, in kelvins.
COPC = 270/(300 − 270)
COPC
=
270
30
COPC = 9
Step 4 · Find share of the Carnot COP (r)
Use r = 100 × COP/COPC
How close the fridge comes to the best possible.
r
=
100
×
3
9
r
=
33.3%
Check
Put the numbers back in. Both sides should match.
1,200 = 900 + 300 ✓
3
=
900
300
✓
9 = 270/(300 − 270) ✓
33.3
≈
100
×
3
9
✓
Topic check
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