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define-efficiency-of-heat-engines

🚀 The second law of thermodynamics states that the total entropy of an isolated system can never decrease over time. This law implies that energy transformations are not 100% efficient, and some energy is always lost as waste heat. The efficiency of heat engines is a measure of how well they convert heat energy into work. It is defined as the ratio of the work output of the engine to the heat input, expressed as a percentage. The second law also introduces the concept of irreversibility in natural processes, indicating that energy tends to disperse or spread out unless constrained.

Theory Explanation

Understanding Heat Engines

A heat engine operates by absorbing heat from a high-temperature source, converting some of that heat into work, and releasing the remaining heat to a low-temperature sink. The efficiency of a heat engine is determined by how much of the absorbed heat is converted into useful work.

\[ \eta = \frac{W}{Q_{in}} \]
Calculating Efficiency

The efficiency (\eta) of a heat engine can be calculated using the formula: \eta = \frac{W}{Q_{in}} \times 100\% where W is the work done by the engine and Q_{in} is the heat absorbed from the hot reservoir.

\[ \eta = \frac{W}{Q_{in}} \times 100\% \]

Key Points

  • 🎯 The second law of thermodynamics states that energy transformations are not 100% efficient.
  • 🎯 Efficiency is defined as the ratio of useful work output to heat input.
  • 🎯 Heat engines convert heat energy into work, but some energy is always lost as waste heat.
  • 🎯 The maximum efficiency of a heat engine is determined by the temperatures of the hot and cold reservoirs.
  • 🎯 Real engines cannot achieve the maximum efficiency due to irreversibilities.

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Examples:💡

Example 1: A heat engine absorbs 500 J of heat from a hot reservoir and does 200 J of work. What is its efficiency?

Solution:

Step 1: Identify the heat input (Q_in) and work output (W). Here, Q_in = 500 J and W = 200 J.

\[ Q_{in} = 500 \text{ J}, W = 200 \text{ J} \]

Step 2: Use the efficiency formula: \eta = \frac{W}{Q_{in}} \times 100\%.

\[ \eta = \frac{200}{500} \times 100\% = 40\% \]

Example 2: A heat engine operates between a hot reservoir at 600 K and a cold reservoir at 300 K. Calculate the maximum efficiency of this engine.

Solution:

Step 1: Use the Carnot efficiency formula: \eta_{max} = 1 - \frac{T_{cold}}{T_{hot}}.

\[ \eta_{max} = 1 - \frac{300}{600} = 0.5 \]

Step 2: Convert to percentage: \eta_{max} = 0.5 \times 100\% = 50\%.

\[ \eta_{max} = 50\% \]

Common Mistakes

  • Mistake: Confusing heat input with work output when calculating efficiency.

    Correction: Always ensure to identify Q_in as the heat absorbed and W as the work done.

  • Mistake: Assuming that all heat input is converted to work.

    Correction: Remember that some energy is always lost as waste heat, and efficiency will always be less than 100%.

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