Physics · Thermodynamics · NEET
For eta to equal 1, the fraction T2/T1 must be 0. A fraction is zero only when its top (T2) is zero. So the sink would need T2 = 0 K, which is absolute zero. Absolute zero cannot be reached, so T2 is always a positive kelvin value. That makes T2/T1 a small positive number, and 1 minus a positive number is always less than 1. Efficiency stays below 100%.
If T1 goes to infinity, T2/T1 goes to 0 and eta goes to 1 in theory. But no real reservoir can have infinite temperature. Real sources like steam or furnace gases have finite T1 (a few hundred to a thousand kelvin). With finite T1 and finite T2, the ratio T2/T1 is always a real positive number, so eta is always below 1.
By the Kelvin-Planck statement of the second law (NCERT): no process can take heat from one reservoir and turn all of it into work with no other change. So the engine must dump some heat Q2 to the sink. Work done W = Q1 - Q2. Since Q2 is never zero, W is always less than Q1, so efficiency eta = W/Q1 is always less than 1.
The first law (energy conservation) alone would allow 100% efficiency - it only says energy is conserved, not the direction of flow. NCERT notes many processes obey the first law yet never happen. The second law is the extra rule that forbids full conversion of heat into work. So 100% efficiency does not break energy conservation; it breaks the second law.
The formula eta = 1 - T2/T1 gives a fraction (a number between 0 and 1). To write it as a percentage, multiply by 100. For example, if T2/T1 = 0.732, then eta = 0.268 = 26.8%. Always put T1 and T2 in kelvin before dividing, never in Celsius.
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
Try the real previous-year questions from this chapter — each with the answer and a full solution.
Because eta = 1 - T2/T1 equals 1 only if T2 = 0 K (absolute zero) or T1 = infinity. Neither is physically possible, so T2/T1 is always a positive number and efficiency stays below 1.
The second law of thermodynamics, in the Kelvin-Planck form: no process can convert all the heat absorbed from a reservoir into work with no other change. The first law (energy conservation) does not forbid it.
No. Carnot's theorem (NCERT) states no engine working between the same two temperatures can have efficiency greater than the Carnot engine. Real engines are less efficient due to friction and irreversibility.
If T2 = 0 K, then eta = 1 - 0/T1 = 1 = 100%. But absolute zero cannot be reached, so this is only a theoretical limit, never achieved.
Both increase efficiency because they lower the ratio T2/T1. Raising T1 or lowering T2 pushes eta closer to 1, but it can never reach 1 for finite T1 and T2 above 0 K.
