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?

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

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.