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Refrigerators & the second law University Physics III · Thermodynamics

Entropy Ferry

Supply work to a reversible heat pump while meeting cooling and cargo-temperature limits.

Best score0 / 300 0 / 9 stars
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Challenge 1 of 3

Cold-store crossing

A reversible refrigerator removes heat from a 250 K reservoir and rejects it to a 300 K reservoir.

Mission targetHeat removed from cold reservoir950–1,050 J
SETUP PREVIEWCOLD-STOREHOT RESERVOIR300 KCOLD RESERVOIR250 KCARNOTrefrigeratorQₕ = 300.0 JQcold = 250.0 JW = 50.0 JCOP 5.00Qhot = Qcold + W · Ideal reversible operation; arrows show energy transfer.

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Cold / hot reservoirs250 K / 300 K
Modelreversible Carnot refrigerator
Energy amountsjoules per cycle

Remove 950–1050 J from the cold reservoir in one cycle.

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Test 1

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50 J
50 J400 J

3 stars on your first test; 2 within three; 1 after that. A first clear earns 100 XP plus 10 per star. Better star records earn the difference; repeating a record earns no extra XP.

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Use a Carnot refrigerator's coefficient of performance and energy balance to move a required amount of heat from a cold reservoir.

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The physics behind the game

Use a Carnot refrigerator's coefficient of performance and energy balance to move a required amount of heat from a cold reservoir.

COPᵣ = T꜀/(Tₕ − T꜀); Q꜀ = COPᵣ W; Qₕ = Q꜀ + W

Model assumptions. A reversible refrigerator transfers heat between two ideal constant-temperature reservoirs, with 0 < T꜀ < Tₕ in kelvin. Its Carnot COP is an upper bound; real refrigerators need more work for the same cooling duty. Work and heat are energy amounts in joules, not powers.