Exams › NEET › Physics

An ideal gas heat engine operates in a Carnot cycle between 227°C and 127°C. It absorbs 6 kcal at the higher temperature. The amount of heat (in kcal) converted into work is equal to

1. 1.2
2. 4.8
3. 3.5
4. 1.6

Correct answer: 4.8

Solution

The efficiency of a Carnot engine is given by η = 1 - (T2/T1), where T1 and T2 are the absolute temperatures of the hot and cold reservoirs, respectively. Converting the given temperatures to Kelvin, T1 = 227 + 273 = 500 K and T2 = 127 + 273 = 400 K. Thus, η = 1 - (400/500) = 0.2. The work done is η × Q1 = 0.2 × 6 = 1.2 kcal.

Related NEET Physics questions

• If for a gas \( \frac{\boldsymbol{R}}{\boldsymbol{C}_{\boldsymbol{v}}}=\mathbf{0 . 6 7}, \) then the gas is made up of molecules which are :
• Use of thermometer is based on which law of thermodynamics?
• The total kinetic energy of 1 mole of \( N_{2} \) at \( 27 \mathrm{C} \) will be approximately
• The quantity \( \frac{2 U}{f k T} \) represents (where \( U= \) internal energy of gas
• A cyclic heat engine does \( 50 \mathrm{kJ} \) of work per cycle. If efficiency of engine is \( 75 \% \) the heat rejected per cycle will be:
• Thermistor material is pressed

⚔️ Practice NEET Physics free + battle 1v1 →