Chapter 20: Problem 18
20.18. A Carnot device extracts 5.00 \(\mathrm{kJ}\) of heat from a body at \(-10.0^{\circ} \mathrm{C} .\) How much work is done if the device exhausts heat into the environment at \((a) 25.0^{\circ} \mathrm{C} ;(\mathrm{b}) 0.0^{\circ} \mathrm{C} ;(\mathrm{c})-25.0^{\circ} \mathrm{C} ;\) In each case, is the device acting as an engine or as a refrigerator?
Short Answer
Step by step solution
Convert Temperatures to Kelvin
Calculate Efficiency of Carnot Device
Case a: Efficiency and Work at 25.0°C
Case b: Efficiency and Work at 0.0°C
Case c: Efficiency and Work at -25.0°C
Conclusion
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Thermodynamics
Some important concepts in thermodynamics include:
- The First Law of Thermodynamics: Energy cannot be created or destroyed, only transferred or transformed.
- The Second Law of Thermodynamics: Heat energy will naturally flow from a hot body to a cold one.
Understanding thermodynamics is crucial for designing efficient engines, refrigerators, and other systems that rely on heat transfer.
Heat Engine
The Carnot engine is an idealized example of a heat engine where all processes are reversible and there are no energy losses. In your exercise, when the Carnot device operates as a heat engine, it would absorb heat (like the 5.00 kJ here), convert a portion into work, and exhaust the remainder to a cold reservoir.
Key points about a heat engine include:
- It follows a cycle, continuously converting heat into work.
- Its efficiency depends on the temperature difference between the hot and cold reservoirs.
Refrigerator
In the context of the Carnot cycle, when the given device acts as a refrigerator, as in cases (a) and (b) of your exercise, it extracts heat from the colder environment and expels it to a warmer one. This is crucial in cooling applications, like domestic fridges and air conditioners.
Key characteristics of a refrigerator include:
- It requires an external work input to move heat against the natural flow (from cold to hot).
- Its performance is often measured by the Coefficient of Performance (CoP), which is different from efficiency.
Efficiency
- \(T_c\) is the absolute temperature of the cold reservoir.
- \(T_h\) is the absolute temperature of the hot reservoir.
Real-world systems always have efficiencies less than a Carnot engine due to losses and irreversibilities. However, understanding this theoretical limit helps engineers strive for designs that approach it, optimizing the balance of work done and energy used.