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A heat engine having efficiency of 70% is used to drive a refrigerator having a coefficient of performance of 5?
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A heat engine having efficiency of 70% is used to drive a refrigerator...
Explanation:

When a heat engine and a refrigerator are connected in series, the overall efficiency can be calculated using the following formula:

Overall Efficiency = (Efficiency of Heat Engine) * (Coefficient of Performance of Refrigerator) / (Efficiency of Heat Engine + Coefficient of Performance of Refrigerator - 1)

Given that the efficiency of the heat engine is 70% and the coefficient of performance of the refrigerator is 5, we can substitute these values into the formula to get:

Overall Efficiency = (0.7) * (5) / (0.7 + 5 - 1) = 0.53 or 53%

This means that only 53% of the heat input to the system is converted into useful work, while the rest is rejected to the environment as waste heat.

Interpretation:

This result highlights the limitations of using a heat engine to drive a refrigerator. While the refrigerator is able to extract useful work from the system, the overall efficiency is relatively low due to the inefficiencies of the heat engine. This suggests that alternative methods of refrigeration may be more efficient, such as using a thermoelectric cooler or a vapor compression cycle.

Conclusion:

In conclusion, the efficiency of a heat engine and the coefficient of performance of a refrigerator must be considered when designing a system that uses both. While this combination can be effective for certain applications, the overall efficiency may be limited by the inefficiencies of the heat engine. Therefore, other methods of refrigeration may be more suitable for certain situations.
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A heat engine having efficiency of 70% is used to drive a refrigerator having a coefficient of performance of 5?
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