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An engine transfers 2.00×10³ J of energy from a hot reservoir during a cycle and transfers 1.50×10³ J to a cold reservoir, what's the efficiency of the engine

a. 0.250
b. 0.500
c. 0.150
d. 0.750

User Virgo
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1 Answer

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Final answer:

The engine's efficiency is found by dividing the work done (the difference between the heat taken from the hot reservoir and the heat given to the cold reservoir) by the heat input. It equates to 25%, making the correct answer choice (a).

Step-by-step explanation:

The efficiency of an engine is calculated by dividing the work output by the heat input and expressing it as a percentage. In this case, the engine transfers 2.00×10³ J of energy from a hot reservoir and 1.50×10³ J to a cold reservoir. The work done by the engine, which is the energy output, is the difference between the energy taken from the hot reservoir and the energy given to the cold reservoir. Mathematically, efficiency (e) can be represented as:

e = (Work done by the engine) / (Energy taken from the hot reservoir)

e = (2.00×10³ J - 1.50×10³ J) / 2.00×10³ J

e = 0.50×10³ J / 2.00×10³ J

e = 0.25 or 25%

The efficiency of the engine is 0.250 or 25%, which corresponds to answer choice (a).

User Jim Mischel
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