A heat engine with an efficiency of 25% takes 1400 J of heat from the high temperature heat source and does J of work in each cycle.
Question
A heat engine with an efficiency of 25% takes 1400 J of heat from the high temperature heat source and does J of work in each cycle.
Solution
It seems like you forgot to mention the amount of work done in each cycle. However, I can still explain how to calculate it based on the information given.
The efficiency of a heat engine is defined as the ratio of the work done to the heat absorbed from the high temperature source. Mathematically, this can be expressed as:
Efficiency = Work done / Heat absorbed
Given that the efficiency is 25% (or 0.25 when expressed as a decimal) and the heat absorbed is 1400 J, we can rearrange the formula to solve for the work done:
Work done = Efficiency * Heat absorbed
Substituting the given values:
Work done = 0.25 * 1400 J = 350 J
So, the heat engine does 350 J of work in each cycle.
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