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Answer :
Final answer:
A 30% efficient heat engine that wastes 1200 J of energy can do approximately 514.29 J of work, but since this exact value is not an answer choice and must be rounded to the nearest hundred, a) 400 J is the closest option, though it is not the precise calculated value. So the closest correct answer is Option A.
Explanation:
The efficiency of a heat engine is calculated by dividing the work output (work done by the engine) by the heat input and then multiplying by 100 to get a percent. Since the engine is 30% efficient, for every 100 J of heat input, the engine converts 30 J to work.
The engine wastes 1200 J, which represents the heat energy that is not converted to work; this is the heat discharged to the environment. To find the heat input, we need to determine what amount of energy results in 1200 J being wasted with a 70% waste rate (100% - 30% efficiency = 70% waste). The equation for heat input (Q_in) with respect to waste energy (Q_out) and efficiency (η) is Q_in = Q_out / (1 - η).
Plug in our values: Q_in = 1200 J / (1 - 0.30) = 1200 J / 0.70 = 1714.29 J. This means that for the heat engine to waste 1200 J, 1714.29 J must be input into the system. Now we calculate the work (W) done by the engine using the efficiency: W = η * Q_in = 0.30 * 1714.29 J = 514.29 J. The engine can do approximately 514 J of work in that one minute, but since this is not one of the given answer choices and we need a whole number, we approximate to the nearest hundred, which is 500 J. Therefore, option a) 400 J is incorrect, but it's the closest to our calculated work output from the provided options.
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