Physics · Thermodynamics · NEET
Efficiency is always W divided by Q1, the heat you take IN from the source. eta = W/Q1. Think of it as: useful output divided by what you paid for. You paid for Q1 (the heat from the source). Q2 is the waste heat thrown to the sink, so it never goes in the bottom of the efficiency fraction. Since W = Q1 - Q2, you can also write eta = (Q1 - Q2)/Q1 = 1 - Q2/Q1.
Because Q2 (heat dumped to the sink) can never be zero for a real cyclic engine. The Second Law of Thermodynamics forbids converting all absorbed heat into work in a cycle. So Q2 is always positive, which makes eta = 1 - Q2/Q1 always less than 1. In practice eta is well below 1 (a Carnot engine between 373 K and 273 K gives only about 26.8%).
Q1 is the heat ABSORBED from the hot source (input). Q2 is the heat REJECTED to the cold sink (waste). Both are taken as positive magnitudes here. The engine converts the difference into work: W = Q1 - Q2. If you swap them, your answer will be wrong, so remember: Q1 is bigger, Q2 is smaller, and the gap between them becomes work.
Always Kelvin when using the temperature form eta = 1 - T2/T1 (this form is for an ideal or Carnot engine). Convert first: T(K) = T(C) + 273. Using Celsius directly is the most common NEET mistake. Note: the general formula eta = 1 - Q2/Q1 uses heats and needs no temperature conversion; only the ideal-engine temperature form needs Kelvin.
A working substance (usually a gas) runs in a repeating cycle: it takes heat from the source, expands and pushes a piston to do work, then rejects leftover heat to the sink and returns to its start state. Because it is a cycle, the internal energy change over one full cycle is zero, so by the First Law the net work equals net heat: W = Q1 - Q2.
The efficiency of an ideal heat engine working between the freezing point and boiling point of water is:
A Carnot engine has an efficiency of 50% when its source is at 327 C. The temperature of the sink is:
A Carnot engine having an efficiency of 1/10 as a heat engine is used as a refrigerator. If the work done on the system is 10 J, the amount of energy absorbed from the reservoir at lower temperature is:
Try the real previous-year questions from this chapter — each with the answer and a full solution.
It absorbs heat Q1 from a hot source, converts part of it into work W, and rejects the rest Q2 to a cold sink, repeating in a cycle.
eta = W/Q1 = 1 - Q2/Q1. For an ideal (Carnot) engine it also equals 1 - T2/T1 with temperatures in Kelvin.
No. Since Q2 is always positive for a cyclic engine, eta = 1 - Q2/Q1 is always less than 1. Efficiency of 100% is impossible by the Second Law.
W (the useful work output) is the numerator and Q1 (heat taken from the source) is the denominator: eta = W/Q1.
To reject leftover heat Q2 and return the working substance to its starting state so the cycle can repeat. Without a sink there is no cycle and no continuous work.
