❄️ Thermodynamics – Stage 2

Page 8 : Refrigerator & Heat Pump (JEE Advanced)

Refrigerator & Heat Pump questions are conceptual + numerical traps in JEE Advanced. One wrong interpretation → full question gone.

1️⃣ What is a Refrigerator?

A refrigerator is a device that removes heat from a low-temperature body and rejects it to a high-temperature body by doing external work.
✔ Heat flows from cold to hot only when work is supplied.

2️⃣ Energy Flow in a Refrigerator

Heat extracted from cold body = Q₂
Work done on system = W
Heat rejected to hot body = Q₁
Energy balance:
Q₁ = Q₂ + W

3️⃣ Coefficient of Performance (COP)

COP of Refrigerator = Heat extracted / Work done
COPR = Q₂ / W
⚠ COP is NOT efficiency
⚠ COP can be greater than 1

4️⃣ What is a Heat Pump?

A heat pump is a device that supplies heat to a hot body by extracting heat from a cold body using work.
👉 Purpose: Heating (not cooling)

5️⃣ COP of Heat Pump

COP of Heat Pump = Heat delivered / Work done
COPHP = Q₁ / W

6️⃣ Relation Between Refrigerator & Heat Pump

Same machine → different objective
Important Relation:
COPHP = COPR + 1

7️⃣ Carnot Refrigerator (Ideal Case)

Maximum possible COP is given by Carnot cycle.
COPR,max = T₂ / (T₁ − T₂)
COPHP,max = T₁ / (T₁ − T₂)
⚠ Temperatures must be in KELVIN

8️⃣ JEE Advanced Traps

  • Confusing Q₁ and Q₂
  • Using °C instead of K
  • Calling COP as efficiency
  • Wrong sign of work

9️⃣ Ranker’s Memory Box

Refrigerator → Q₂ is useful
Heat Pump → Q₁ is useful
COP can be > 1
Always convert °C → K

✅ Stage 2 – Page 8 Completed
➡️ Next: Stage 2 – Page 9 : Carnot Engine & Absolute Temperature Scale

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