In one cycle a heat engine absorbs 530 J from a high-temperature reservoir and expels 310 J to a low-temperature reservoir. If the efficiency of this engine is 55% of the efficiency of a Carnot engine, what is the ratio of the low temperature to the high temperature in the Carnot engine?
The work out of the heat engine is 530-310 = 220 J per cycle and its efficiency is 220/530 = 41.5%
If the Carnot engine is 1/0.55 times higher than 41.5%, that would be 75.5%
The Tlow/Thi ratio of a Carnot engine with 74.5% effiency would have to be
1 - Tlow/Thi = 0.745
Tlow/Thi = 0.255
To solve this problem, we can use the formula for the efficiency of a heat engine:
Efficiency = (1 - Tlow / Thigh) * 100%
Let's assume the efficiency of the Carnot engine is x, then the efficiency of the given engine is 0.55x.
We are given that the given engine absorbs 530 J and expels 310 J. We can calculate the ratio of heat absorption to heat expulsion:
Heat Absorption / Heat Expulsion = 530 / 310
Now, we can set up the equation for the efficiency of the given engine:
0.55x = (1 - Tlow / Thigh) * 100%
To simplify the equation, let's divide both sides by 100:
0.0055x = 1 - Tlow / Thigh
Now, we can substitute the ratio of heat absorption to heat expulsion:
0.0055x = 1 - (310 / 530)
Simplifying further:
0.0055x = 1 - 0.5849
0.0055x = 0.4151
Now, let's isolate x:
x = 0.4151 / 0.0055
x = 75.56
So, the efficiency of the Carnot engine is approximately 75.56%.
Now, let's use this efficiency to find the ratio of the low temperature (Tlow) to the high temperature (Thigh) in the Carnot engine:
Efficiency = (1 - Tlow / Thigh) * 100%
75.56% = (1 - Tlow / Thigh) * 100%
Dividing both sides by 100:
0.7556 = 1 - Tlow / Thigh
Rewriting the equation:
Tlow / Thigh = 1 - 0.7556
Tlow / Thigh = 0.2444
So, the ratio of the low temperature to the high temperature in the Carnot engine is approximately 0.2444.
To solve this problem, we need to understand the principles of heat engines.
A heat engine is a device that converts thermal energy (heat) into mechanical work. The efficiency of a heat engine is defined as the ratio of the work output to the heat input.
In this particular problem, the efficiency of the given engine is said to be 55% of the efficiency of a Carnot engine. The Carnot engine is an idealized heat engine that operates between two heat reservoirs at different temperatures.
Let's assume that the high-temperature reservoir in the Carnot engine has a temperature of Th (in Kelvin), and the low-temperature reservoir has a temperature of Tl (in Kelvin).
The efficiency of a Carnot engine is given by the formula:
Efficiency (Carnot) = 1 - Tl / Th
We are also given that the given engine absorbs 530 J from the high-temperature reservoir and expels 310 J to the low-temperature reservoir.
Using the formula for efficiency, we can set up the following equation for the given heat engine:
Efficiency (Given) = (Work Output) / (Heat Input)
= (530 J - 310 J) / 530 J
= 220 J / 530 J
Since the given engine's efficiency is said to be 55% of the efficiency of a Carnot engine, we can set up the following equation:
Efficiency (Given) = 0.55 * Efficiency (Carnot)
Substituting the values, we have:
220 J / 530 J = 0.55 * (1 - Tl / Th)
Simplifying the equation, we get:
1 - Tl / Th = (220 J / 530 J) / 0.55
Now, we can solve this equation to find the ratio of low temperature to high temperature (Tl / Th).
Tl / Th = 1 - ((220 J / 530 J) / 0.55)
Using the given values and performing the calculation, we will find the ratio of low temperature to high temperature in the Carnot engine.