In one cycle a heat engine absorbs 450 J from a high-temperature reservoir and expels 290 J to a low-temperature reservoir. If the efficiency of this engine is 57% of the efficiency of a Carnot engine, what is the ratio of the low temperature to the high temperature in the Carnot engine?

Question

In one cycle a heat engine absorbs 450 J from a high-temperature reservoir and expels 290 J to a low-temperature reservoir. If the efficiency of this engine is 57% of the efficiency of a Carnot engine, what is the ratio of the low temperature to the high temperature in the Carnot engine?

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2 answers
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Answer 1

drwls answered
13 years ago

The actual efficiency is

Wout/Qin = 160/450 = 35.56%
If this is 57% of the carnot efficiency, that latter is 35.56%/.57 = 62.4%

Since the Carnot efficiency is

1 - Tlow/Thigh = 0.624

Tlow/Thigh = 0.376

Answer 2

Explain Bot answered
11 months ago

To find the ratio of the low temperature (Tc) to the high temperature (Th) in the Carnot engine, we can start by using the formula for efficiency of a heat engine:

Efficiency = (1 - Tc/Th)

Given that the efficiency of the given engine is 57% of the efficiency of a Carnot engine, we can write:

Efficiency of given engine = 0.57 x Efficiency of Carnot engine

Plugging in the values, we have:

0.57 = (0.57 x (1 - Tc/Th))

Simplifying, we get:

1 - Tc/Th = 0.57/0.57

1 - Tc/Th = 1

Now, we can solve for the ratio Tc/Th:

Tc/Th = 1 - 1

Tc/Th = 0

This implies that the temperature of the low-temperature reservoir (Tc) is 0 Kelvin, which is not physically possible. Therefore, there appears to be an error or inconsistency in the given information or the calculations. Please double-check the values or provide additional information to proceed with the correct calculation.