Calculating Work Output from Thermal Efficiency of Car Engine

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SUMMARY

The discussion focuses on calculating the work output of a car engine with a power output of 600 kilowatts per hour and a thermal efficiency of 30.0%. The key equation used is thermal efficiency = w(out)/Q(hot). The participant attempted to convert kilowatts to kilojoules per second and adjust for efficiency but encountered confusion regarding the calculations. It was noted that the 600 kW/hr represents the power output after a significant energy loss, indicating the need for precise conversion and understanding of thermal efficiency in the context of engine performance.

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  • Understanding of thermal efficiency in thermodynamics
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Homework Statement


The power output of a car engine running at 1900 is 600 kilowatts per hour.
How much work is done per cycle if the engine's thermal efficiency is 30.0 percent

Homework Equations



thermal efficiency=w(out)/Q(hot)

The Attempt at a Solution


It wants the answer in KJ/s so i converted the kilowatts to KJ/sec then divided by rev/s then multiplied by .3. What am i doing wrong? Any help would be appreciated.
 
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Looks like you should be multiplying by 10/3.
The 600 kw/hr seems to be the power output after 70% of the initial energy has been lost.
 

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