The Carnot engine is an ideal engine with maximum efficiency.
Chapter 11, “Thermodynamics,” is essential for the 2024-2025 CBSE Class 11 Physics exams. It focuses on MCQs that explore key concepts like the laws of thermodynamics, entropy and heat engines. These concepts are crucial for understanding how energy is transferred and converted. The MCQs will test your knowledge of energy, efficiency and heat transfer in different situations. To succeed, focus on understanding the fundamental principles and practice regularly. Consistent revision will help reinforce these concepts. With thorough preparation and problem-solving practice, you’ll be confident in tackling thermodynamics-related questions in your exams.
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The efficiency of a Carnot engine depends on the temperatures of the heat source and the heat sink. It is expressed mathematically as:
η = 1 – (T_c / T_h)
Where:
– η is the efficiency of the Carnot engine (in decimal form).
– T_c is the absolute temperature of the cold reservoir (heat sink), measured in kelvins (K).
– T_h is the absolute temperature of the hot reservoir (heat source), also measured in kelvins (K).
### Key Points:
1. Absolute Temperatures: The temperatures T_c and T_h must be in absolute units (kelvins) for the formula to be valid. This ensures consistency and proper calculation of the efficiency.
2. Dependence on Temperature Difference: The efficiency increases as the temperature difference between the hot and cold reservoirs increases. Specifically:
– Higher T_h (hot reservoir temperature) leads to higher efficiency.
– Lower T_c (cold reservoir temperature) also leads to higher efficiency.
3. Maximum Theoretical Efficiency: The Carnot engine represents an idealized heat engine with the maximum possible efficiency dictated by thermodynamics. No real engine can surpass this efficiency due to practical limitations such as friction, heat losses, and irreversibilities.
4. Efficiency Limit: The efficiency is always less than 1 (or less than 100%) because T_c is always greater than 0 K, and T_h is finite in practical scenarios.
Example Calculation:
Suppose a Carnot engine operates with a heat source at T_h = 500 K and a heat sink at T_c = 300 K.
Efficiency, η = 1 – (T_c / T_h)
= 1 – (300 / 500)
= 1 – 0.6
= 0.4 or 40%
Thus, the Carnot engine’s efficiency in this case is 40%.