# Heat Efficiency of a Carnot Engine

• castrodisastro
In summary, the efficiency of the Carnot engine is increased by a factor of 5.00 when the temperature of the warmer reservoir is increased by a factor of 2.00.
castrodisastro

## Homework Statement

A Carnot engine operates between a warmer reservoir at a temperature T1 and a cooler reservoir at a temperature T2. It is found that increasing the temperature of the warmer reservoir by a factor of 2.00 while keeping the same temperature for the cooler reservoir increases the efficiency of the Carnot engine by a factor of 5.00. Find the efficiency of the engine and the ratio of the temperatures of the two reservoirs in their original form.

ε=1-(T2/T1)

## The Attempt at a Solution

I think we are supposed to get a numerical value for an answer but I don't think I have enough information.

ε=1-(T2/T1)

The question states, "increasing the temperature of the warmer reservoir by a factor of 2.00 while keeping the same temperature for the cooler reservoir increases the efficiency of the Carnot engine by a factor of 5.00"

That means T1 will be doubled and once that happens, ε is quintupled, resulting in...

5ε=1-(T2/2T1)

The question asks for the efficiency, which I am supposed to answer with a numerical value. So I tried solving for ε.

ε=(1/5)-(T2/10T1))

I can't think of a way to solve this with 2 independent variables. Am I ignoring something from the question that would point me towards a value? Maybe I'm forgetting some mathematical property?

Once I have the efficiency then I can solve for the ratio of the two temperatures. Since it asks me for the ratio, I assume that knowing T1 and T2 are not required, so how could we solve for ε and get a numerical value?

Use R for ##T_1/T_2##. Now you have 2 equations in 2 unknowns!

1 person
BvU said:
Use R for ##T_1/T_2##. Now you have 2 equations in 2 unknowns!

Thanks! I knew it was staring at me in the face.

## 1. What is a Carnot engine?

A Carnot engine is a theoretical heat engine that operates on the Carnot cycle, which is a reversible cycle that consists of two isothermal (constant temperature) and two adiabatic (no heat transfer) processes. It serves as a model for the most efficient possible engine that can be built with a given pair of heat reservoirs.

## 2. How is the efficiency of a Carnot engine calculated?

The efficiency of a Carnot engine is calculated by dividing the temperature difference between the hot and cold reservoirs by the temperature of the hot reservoir. This can be expressed as: efficiency = (Th - Tc) / Th, where Th is the temperature of the hot reservoir and Tc is the temperature of the cold reservoir.

## 3. What is the maximum theoretical efficiency of a Carnot engine?

The maximum theoretical efficiency of a Carnot engine is 1 - (Tc / Th), where Tc is the temperature of the cold reservoir and Th is the temperature of the hot reservoir. This means that the efficiency of a Carnot engine can never reach 100%, as there will always be some heat loss during the cycle.

## 4. What factors affect the efficiency of a Carnot engine?

The efficiency of a Carnot engine is affected by the difference in temperature between the hot and cold reservoirs, as well as the type of working substance used in the engine. Higher temperature differences and more efficient working substances will result in a higher efficiency.

## 5. Can a Carnot engine ever reach 100% efficiency?

No, a Carnot engine can never reach 100% efficiency in practice due to factors such as friction and heat loss. However, it provides a benchmark for the maximum possible efficiency that can be achieved by any heat engine operating between two given temperatures.

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