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AC with 50% of Carnot COP: max surrounding temperature

Short answer

COP_max = T_C/ΔT = 293.15/ΔT; actual COP = 0.5×293.15/ΔT. Cooling load Q_L = 0.8 ΔT kW = COP × 1.8 kW → 0.8ΔT = (146.6/ΔT)×1.8 → ΔT²=329.8, ΔT=18.2 K. Maximum surrounding temperature = 293.15+18.2 = 311.35 K ≈ 38.2°C.

An air conditioning unit having COP 50% of the theoretical maximum maintains a house at a temperature of 20°C by cooling it against the surrounding temperature. The house gains energy at a rate of 0.8 kW per degree temperature difference. For a maximum work input of 1.8 kW, determine the maximum surrounding temperature for which it provides sufficient cooling.

Given: house temperature T_C = 20 °C = 293.15 K, COP = 0.5 × COP_max, heat gain of 0.8 kW per °C of temperature difference, W_max = 1.8 kW.

Let T_H = outside (surrounding) temperature in K, and ΔT = T_H − T_C.

Step 1: COP of the unit

COP_max (Carnot) = T_C/(T_H − T_C) = 293.15/ΔT COP_actual = 0.5 × 293.15/ΔT = 146.6/ΔT

Step 2: Heat the unit must remove

The house gains heat at 0.8 ΔT kW, so to keep the temperature steady the cooling rate must be

Q_L = 0.8 ΔT kW

Step 3: Use the power limit

Q_L = COP × W

0.8 ΔT = (146.6/ΔT) × 1.8

ΔT² = 146.6 × 1.8/0.8 = 329.8

ΔT = 18.16 K

Step 4: Maximum surrounding temperature

T_H = 293.15 + 18.16 = 311.31 K = 38.2 °C

Answer: the unit gives sufficient cooling up to a maximum surrounding temperature of about 38.2 °C (ΔT ≈ 18.2 K).

Check: Q_L = 0.8 × 18.16 = 14.53 kW, COP = 8.071, W = 1.80 kW ✓.

✓ Answer checked by Claude Opus 5.5 (high effort)

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