Refrigerators and Heat Pumps: COP, Energy Balance, Why Rooms Warm

Concept: refrigeration as heat pumping. A refrigerator uses external work to force heat to move from the cold interior to the warmer surroundings. This is the reverse of the spontaneous heat-flow direction, so work is required.

Concept: second law allows cold→hot with work input. Moving heat from cold to hot is not spontaneous, but it’s allowed if you supply work. A refrigerator doesn’t violate the second law because it consumes electrical energy.

Concept: energy conservation for refrigerators. The heat dumped to the room equals the heat removed from inside plus the work input. That’s why q_h is always larger than q_c when w>0.

Concept: using q_h=q_c+w. Add the heat removed and the work input to find heat released. q_h=400+150=550 j.

Concept: net heat added to the room equals w. With the door open, the fridge still dumps q_h into the room while removing q_c from the same room air. Since q_h=q_c+w, the net effect is adding w as heat, warming the room overall.

Concept: room gains the work input. The fridge releases q_h=q_c+w to the room but 'takes' q_c from the interior, so those cancel. The remaining net gain to the room is w=200 j.

Concept: COP measures cooling per work. Refrigerator COP compares how much heat is removed from the cold space to how much work is required. So COP=q_c/w.

Concept: meaning of high COP. A higher COP means you remove more heat from the cold region for the same work input. That indicates better performance (more cooling per joule of electricity).

Concept: COP calculation. COP=q_c/w. Here COP=900/300=3.0, meaning 3 joules of heat removed per joule of work input.

Concept: COP is not efficiency. COP compares heat moved to work input, so it can exceed 1 because the device is transporting heat rather than creating energy. This does not violate conservation of energy or the second law.

Concept: heat pump heating performance. For heating, you care about how much heat is delivered to the house per work input. That ratio is COP_h=q_h/w.

Concept: COP_h calculation. COP_h=q_h/w. Substituting gives 2400/600=4.0, meaning 4 joules of heat delivered per joule of work.

Concept: heat pumps move heat. A heat pump brings in additional heat from outside and adds it to the house along with the electrical work input. Therefore q_h can be larger than w without violating any laws.

Concept: energy balance and COP meaning. Refrigerators and heat pumps both satisfy q_h=q_c+w, so expelled heat exceeds removed heat when w>0. COP is not constrained to be less than 1 because it is a performance ratio, not an efficiency.

Concept: heat rejection at the condenser coil. The warm back coil is where the refrigerator dumps heat into the room. That heat includes both the heat removed from the inside and the electrical work input.

Concept: cold→hot requires work. Without work input, heat would be flowing from cold to hot spontaneously. That contradicts the second law, so w=0 would make the device impossible.

Concept: solving w from q_h=q_c+w. Rearrange to w=q_h−q_c. Substituting gives w=650−500=150 j.

Concept: real devices are irreversible. Real refrigerators have friction, electrical resistance, and finite temperature differences, all of which produce entropy. So total entropy (system plus surroundings) increases overall.

Concept: more irreversibility and bigger 'lift' require more work. Losses waste work and increase entropy production, lowering COP. A larger temperature difference makes it harder to move heat, requiring more work per unit heat removed.

Concept: second law for heat pumps. Heat does not spontaneously flow from cold to hot, so work input is required. In real operation, the total entropy of the universe increases because the process is irreversible.