Physics · Thermodynamics · NEET
Always Kelvin. The formula eta = 1 - T2/T1 is only true when both temperatures are absolute (Kelvin). Convert first: T(K) = T(Celsius) + 273. Example: 327 C becomes 327 + 273 = 600 K. If you plug in Celsius, the ratio T2/T1 is wrong and your answer is wrong. This single mistake is the number one reason students lose this mark.
T1 is the SOURCE, the hot reservoir (larger temperature). T2 is the SINK, the cold reservoir (smaller temperature). In the formula eta = 1 - T2/T1, you divide the smaller (sink) by the larger (source). If you ever get an efficiency above 1 or a negative value, you have swapped them.
Rearrange the formula. Start with eta = 1 - T2/T1. Then T2/T1 = 1 - eta, so T2 = T1 (1 - eta). Steps: convert T1 to Kelvin, write efficiency as a decimal (50 percent = 0.50), compute 1 - 0.50 = 0.50, multiply by T1. For T1 = 600 K: T2 = 600 x 0.50 = 300 K = 27 C.
The formula gives a decimal (a fraction between 0 and 1). Multiply by 100 to get percent. Example: 1 - 273/373 = 0.268, so efficiency = 26.8 percent. If the question gives efficiency as a percent, convert it back to a decimal before using it in the formula.
That is impossible for a Carnot engine, so you made an error. The most common cause is putting the sink on top (T1/T2 instead of T2/T1), or forgetting to convert to Kelvin. A correct Carnot efficiency is always between 0 and 1 (0 to 100 percent), and never reaches 100 percent because T2 can never be 0 K.
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 percent 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.
eta = 1 - T2/T1, where T1 is the source (hot) temperature and T2 is the sink (cold) temperature, both in Kelvin. It can also be written as eta = (T1 - T2)/T1.
The formula uses the ratio of absolute temperatures. Only the Kelvin scale starts at absolute zero, so ratios are physically meaningful. Celsius has an arbitrary zero point, so a ratio of Celsius values gives a wrong answer.
No. Efficiency reaches 100 percent only if T2 = 0 K (absolute zero), which is impossible to achieve. So a Carnot engine is always below 100 percent. See why-carnot-efficiency-never-100-percent.
Rearrange to T1 = T2/(1 - eta). Convert the sink temperature to Kelvin first, use efficiency as a decimal, then compute and convert back to Celsius if needed.
No. It depends only on the two reservoir temperatures T1 and T2. That is why every numerical only needs the source and sink temperatures, not the type of gas or the pressure.
