Physics · Thermodynamics · NEET
No. Efficiency of an engine is what you get (work) divided by what you pay (heat), and it is always less than 1. COP is what you want (heat moved) divided by what you pay (work input). Since the heat pulled out or delivered is usually much larger than the small work you supply, COP is normally greater than 1. A COP of 5 just means for every 1 J of electrical work you move 5 J of heat. That does not break energy conservation because Q1 = Q2 + W still holds.
They are the same machine (reverse heat engine); only the useful output differs. In a refrigerator the useful thing is Q2, the heat removed from the cold space, so COP_fridge = Q2/W. In a heat pump the useful thing is Q1, the heat delivered to the warm room, so COP_pump = Q1/W. Because Q1 = Q2 + W, the two are linked: COP_pump = COP_fridge + 1. A heat pump always has COP at least 1.
For a refrigerator, the COLD reservoir temperature goes on top: COP = T2/(T1 - T2). For a heat pump, the HOT reservoir temperature goes on top: COP = T1/(T1 - T2). The denominator (T1 - T2) is the same for both. Remember: the numerator matches the reservoir whose heat you care about.
Always Kelvin. Convert every temperature with T(K) = t(C) + 273 before plugging in. A useful shortcut: the difference (T1 - T2) is the same whether you use Celsius or Kelvin, but the numerator T1 or T2 must be in Kelvin. Forgetting to convert the numerator is the most common mistake.
First find COP from the temperatures: beta = T2/(T1 - T2). Then the work per second (power) is P = Q2/beta, where Q2 is the heat removed per second. So the colder you want the space (small T2) or the bigger the temperature gap, the smaller the COP and the more power you need.
The temperature inside a refrigerator is t2 degree C and the room temperature is t1 degree C. The amount of heat delivered to the room for each joule of electrical energy consumed ideally will be
A refrigerator works between 4 degree C and 30 degree C. It is required to remove 600 calories of heat every second to keep the temperature of the refrigerated space constant. The power required is (Take 1 cal = 4.2 J):
Try the real previous-year questions from this chapter — each with the answer and a full solution.
COP is a measure of how well a refrigerator or heat pump moves heat using a given amount of work. For a refrigerator COP = Q2/W (heat removed per unit work). For a heat pump COP = Q1/W (heat delivered per unit work). It replaces efficiency for reverse-running machines.
The maximum possible COP is that of a Carnot (ideal reversible) refrigerator: COP = T2/(T1 - T2), with temperatures in Kelvin. No real refrigerator can beat this. As T2 approaches T1 the COP grows very large, but it stays finite as long as there is a temperature difference.
No. By the second law of thermodynamics a refrigerator cannot have infinite COP, just as a heat engine cannot have 100% efficiency. Infinite COP would mean moving heat from cold to hot with zero work, which is forbidden.
Since Q1 = Q2 + W, dividing by W gives COP_pump = COP_fridge + 1. So a heat pump's COP is always exactly one more than the refrigerator COP for the same machine, and it is always at least 1.
The Carnot relations Q1/Q2 = T1/T2 hold only for absolute (Kelvin) temperatures. Using Celsius would give wrong ratios. Always convert with T(K) = t(C) + 273 before using T2/(T1 - T2).
