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INTERACTIVE EXPLANATION

How does a fridge move heat out of a cold space?

Look through a cold room and follow its refrigerant loop. Count the heat removed and the work added, then try putting both sides in the same room.

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Make a discovery

A refrigerator moves heat from a colder place to a warmer place using work. The warm side releases the heat removed from the cold side plus the energy supplied as work.

Make a prediction

With both sides in the same isolated room, does this cycle cool that room?

  • Yes, it removes 200 J
  • No, the room gains the supplied work
Read the explanation

The room loses Q_cold but receives Q_hot=Q_cold+W. Its net gain is W, 100 J in the default example.

Understand it

Absorb heat inside

In this commercial cold-room architecture, low-pressure refrigerant absorbs heat in an evaporator. Liquid in a liquid–vapor mixture evaporates, and vapor returns toward the compressor. The refrigerant stays in a closed circuit; it is not the room air.

Compress and reject heat

The compressor receives vapor and raises its pressure. The high-pressure refrigerant rejects heat in the condenser and condenses to liquid. Fans help exchange heat with air but do not send that air through the refrigerant pipe.

Reduce pressure and repeat

A metering valve reduces pressure before the evaporator. Its outlet can be a liquid–vapor mixture. This selected dry-expansion system uses a thermostatic expansion valve; the exact internal valve mechanism is outside this first cutaway.

Look closer at the science

Count energy across a boundary

For one accounting batch, Q_hot=Q_cold+W. The default removes 200 J, adds 100 J of work and delivers 300 J to the warm side. These are chosen example transfers, not a measured appliance cycle.

COP is not a percentage efficiency

Cooling coefficient of performance is Q_cold/W. Heating COP is Q_hot/W, so it is one greater for the same budget. A COP above one is possible because heat is moved as well as work supplied.

An open fridge is no room cooler

If both heat exchangers interact with the same isolated room, the removed heat returns to that room along with work input. The net gain is W. A real controller changes its behavior with the door open; this accounting comparison predicts no time or temperature curve.

Try it yourself: Audit the warm side

Supplies

  • Paper
  • Pencil
  • Six equal energy cards or tally marks
  1. Name the boundaries

    Draw a cold box inside a room. Put two energy cards in the cold box and one outside as supplied work.

  2. Move and add

    Move two cards out of the box. Add the work card and place all three on the warm side. Count before and after.

  3. Include the open box

    Redraw the boundary to include both sides in one room. Two cards leave one part and return to another; one new work card enters the whole room.

Can every energy card find a destination?

Paper energy accounting only. Do not open appliances, handle refrigerants or build mains-powered equipment. This is not a physical COP measurement.

Sources and model limits

  • One qualitative commercial cold-room vapor-compression architecture. It does not claim that every household fridge uses this valve or layout.
  • COP is an imposed accounting assumption. No refrigerant properties, operating pressures, temperature solver, refrigerant mass flow, defrost or controller simulation.
  • Changing the useful-output viewpoint changes accounting only. It does not reverse the physical circuit or simulate a reversing valve.

Cold rooms use a refrigerant circuit linking evaporator and condensing equipment.

Manufacturer system architecture reference, not a universal household-fridge diagram.

Danfoss · Cold-room systems

A thermostatic expansion valve regulates evaporator superheat through a sensing mechanism.

Manufacturer dry-expansion explanation. No quantitative valve model is claimed.

Danfoss · Expansion valves

Independent subject review is pending.

Read the sources and model assumptions