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Ch 20: The Second Law of Thermodynamics
Young & Freedman Calc - University Physics 15th Edition
Young & Freedman Calc15th EditionUniversity PhysicsISBN: 9780135159552Not the one you use?Change textbook
All textbooksYoung & Freedman Calc 15th EditionCh 20: The Second Law of ThermodynamicsProblem 14c
Chapter 20, Problem 14c

A Carnot engine is operated between two heat reservoirs at temperatures of 520520 K and 300300 K. What is the thermal efficiency of the engine?

Verified step by step guidance
1
Understand the concept of a Carnot engine: A Carnot engine is an idealized heat engine that operates between two heat reservoirs and has the maximum possible efficiency. The efficiency of a Carnot engine depends only on the temperatures of the two reservoirs.
Identify the temperatures of the heat reservoirs: In this problem, the high-temperature reservoir (T₁) is at 520 K, and the low-temperature reservoir (T₂) is at 300 K.
Recall the formula for the thermal efficiency of a Carnot engine: The efficiency (η) is given by the equation η = 1 - (T₂/T₁), where T₁ is the temperature of the hot reservoir and T₂ is the temperature of the cold reservoir.
Substitute the given temperatures into the efficiency formula: Replace T₁ with 520 K and T₂ with 300 K in the equation η = 1 - (T₂/T₁).
Simplify the expression: Perform the division and subtraction to find the efficiency in decimal form, which can be converted to a percentage by multiplying by 100.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Carnot Engine

A Carnot engine is a theoretical thermodynamic cycle that is considered the most efficient possible engine. It operates between two heat reservoirs, absorbing heat from the high-temperature reservoir and expelling heat to the low-temperature reservoir, while performing work. The efficiency of a Carnot engine depends solely on the temperatures of these reservoirs.
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Thermal Efficiency

Thermal efficiency is a measure of how well an engine converts heat into work. For a Carnot engine, it is calculated using the formula: efficiency = 1 - (T_cold/T_hot), where T_cold and T_hot are the absolute temperatures of the cold and hot reservoirs, respectively. This formula shows that efficiency increases as the temperature difference between the reservoirs increases.
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Absolute Temperature

Absolute temperature is measured in Kelvin, which is the SI unit for temperature. It starts at absolute zero, the point where all molecular motion ceases. In thermodynamics, using Kelvin is crucial because it allows for direct calculations of efficiency and other properties without negative values, ensuring accurate and meaningful results.
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Related Practice
Textbook Question

A Carnot refrigerator is operated between two heat reservoirs at temperatures of 320320 K and 270270 K. If in each cycle the refrigerator receives 415415 J of heat energy from the reservoir at 270270 K, how many joules of heat energy does it deliver to the reservoir at 320320 K?

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Textbook Question

A certain brand of freezer is advertised to use 730730 kWh of energy per year. What is the theoretical maximum amount of ice this freezer could make in an hour, starting with water at 20.020.0°C?

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Textbook Question

A Carnot engine is operated between two heat reservoirs at temperatures of 520520 K and 300300 K. How much mechanical work is performed by the engine during each cycle?

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Textbook Question

A Carnot engine is operated between two heat reservoirs at temperatures of 520520 K and 300300 K. If the engine receives 6.456.45 kJ of heat energy from the reservoir at 520520 K in each cycle, how many joules per cycle does it discard to the reservoir at 300300 K?

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Textbook Question

A certain brand of freezer is advertised to use 730730 kWh of energy per year. Assuming the freezer operates for 55 hours each day, how much power does it require while operating?

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Textbook Question

A refrigerator has a coefficient of performance of 2.252.25, runs on an input of 135135 W of electrical power, and keeps its inside compartment at 55°C. If you put a dozen 1.01.0-L plastic bottles of water at 3131°C into this refrigerator, how long will it take for them to be cooled down to 5 5°C? (Ignore any heat that leaves the plastic.)

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