The Carnot Cycle: Videos & Practice Problems

The Carnot cycle, developed by Sadi Carnot, represents an ideal reversible heat engine with maximum theoretical efficiency. It consists of four steps: two isothermal (heat absorption and release) and two adiabatic (no heat transfer). The maximum efficiency is given by $1^{\mathrm{-}}\frac{T}{T}$, where T represents the temperatures of the cold and hot reservoirs. Understanding this cycle is crucial for grasping thermodynamics and the limitations of real heat engines.

The Carnot Cycle and Maximum Theoretical Efficiency

The Carnot cycle is a fundamental concept in thermodynamics that illustrates the maximum theoretical efficiency of a heat engine operating between two thermal reservoirs. According to the second law of thermodynamics, no heat engine can achieve 100% efficiency, but the Carnot cycle provides a framework for understanding how close we can get. Developed by Sadi Carnot in the 1800s, this idealized cycle consists of four distinct processes: two isothermal (constant temperature) and two adiabatic (no heat transfer).

In the Carnot cycle, the first step involves isothermal expansion, where the engine absorbs heat \( Q_h \) from the hot reservoir at temperature \( T_h \). This is followed by an adiabatic expansion, where the system does work on the surroundings without heat exchange. The cycle then reverses, with isothermal compression where heat \( Q_c \) is expelled to the cold reservoir at temperature \( T_c \), and finally, an adiabatic compression that returns the system to its initial state.

The efficiency \( \eta \) of a Carnot engine is given by the equation:

\[ \eta = 1 - \frac{T_c}{T_h} \]

This equation indicates that the efficiency depends solely on the temperatures of the hot and cold reservoirs, measured in Kelvin. For example, if \( T_h = 520 \, K \) and \( T_c = 300 \, K \), the maximum theoretical efficiency would be:

\[ \eta = 1 - \frac{300}{520} \approx 0.423 \text{ or } 42.3\% \]

This means that even under ideal conditions, the engine can only convert about 42.3% of the absorbed heat into work.

To calculate the waste heat expelled during the cycle, we can use the relationship between the heat absorbed and the heat expelled, which is expressed as:

\[ \frac{Q_c}{Q_h} = \frac{T_c}{T_h} \]

From this, we can derive \( Q_c \) as follows:

\[ Q_c = Q_h \cdot \frac{T_c}{T_h} \]

For instance, if \( Q_h = 6.45 \, kJ \), substituting the values gives:

\[ Q_c = 6.45 \cdot \frac{300}{520} \approx 3.72 \, kJ \]

Finally, the work done by the engine \( W \) can be calculated using the difference between the heat absorbed and the heat expelled:

\[ W = Q_h - Q_c \]

Using the previous values, we find:

\[ W = 6.45 \, kJ - 3.72 \, kJ \approx 2.73 \, kJ \]

This calculation shows that the maximum work output from the Carnot engine operating under these conditions is approximately 2.73 kJ per cycle. Understanding the Carnot cycle is crucial for grasping the limits of efficiency in real-world heat engines and the principles of thermodynamics.

A theoretical heat engine in space could operate between the Sun's 5500°C surface and the –270.3°C temperature of intergalactic space. What would be its maximum theoretical efficiency?

• A. 99.98%
• B. 95.1%
• C. 99.95%

Answer: (C)

A Carnot engine with an efficiency of 70% is cooled by water at 10°C. What temperature must the hot reservoir be maintained at?

• A. 33.3 K
• B. 404.3 K
• C. 943.3 K
• D. 14.3 K

Answer: (C)

Lifting Mass with a Heat Engine

In this scenario, we are tasked with determining how many cycles a Carnot engine must complete to lift a 10-kilogram mass by 5 meters. To begin, we apply the work-energy principle, which states that the work required to lift an object is given by the formula:

Work (W) = mgh

Here, m is the mass (10 kg), g is the acceleration due to gravity (approximately 9.8 m/s²), and h is the height (5 m). Plugging in these values, we find:

W = 10 kg × 9.8 m/s² × 5 m = 490 joules

This means that 490 joules of work is required to lift the mass. The Carnot engine operates between a hot reservoir and a cold reservoir, performing work by converting heat energy into mechanical energy. The work done by the engine in each cycle can be expressed using the equation:

W = Q_H - Q_C

Where Q_H is the heat absorbed from the hot reservoir and Q_C is the heat rejected to the cold reservoir. In this case, we know that the engine extracts 25 joules of energy from the hot reservoir, so:

Q_H = 25 joules

To find the work done per cycle, we can also use the efficiency of the Carnot engine, which is defined as:

Efficiency (η) = W / Q_H = 1 - (T_C / T_H)

In this case, the temperatures are given as T_H = 182°C (or 455 K) and T_C = 0°C (or 273 K). Substituting these values into the efficiency equation gives:

η = 1 - (273 K / 455 K)

Calculating this yields:

η ≈ 0.4

Now, we can find the work done by the engine per cycle:

W = Q_H × η = 25 joules × 0.4 = 10 joules

With the work done per cycle established, we can now determine the number of cycles required to lift the mass. The total work needed to lift the mass is 490 joules, and since each cycle of the engine does 10 joules of work, we can find the number of cycles (n) using the formula:

n = Total Work / Work per Cycle

Substituting the known values gives:

n = 490 joules / 10 joules = 49 cycles

Thus, it will take 49 cycles of the Carnot engine to lift the 10-kilogram mass by 5 meters. This example illustrates the relationship between energy conservation, work, and the efficiency of heat engines in practical applications.

Your friend claims they have a design for a reversible heat engine that can operate between the freezing and boiling temperatures of water that has an efficiency of 30%. Is this possible?

• A. No
• B. Yes
• C. Not enough information

Answer: (A)

Here’s what students ask on this topic:

What is the Carnot cycle and why is it important in thermodynamics?

The Carnot cycle, developed by Sadi Carnot, is an idealized thermodynamic cycle that represents the most efficient possible heat engine. It consists of four steps: two isothermal processes (heat absorption and release) and two adiabatic processes (no heat transfer). The importance of the Carnot cycle lies in its demonstration of the maximum theoretical efficiency that any heat engine can achieve, which is given by the equation:

$\eta =1-\frac{T_c}{T_h}$

where $T_c$ and $T_h$ are the temperatures of the cold and hot reservoirs, respectively. Understanding the Carnot cycle helps in grasping the limitations of real heat engines and the principles of thermodynamics.

How do you calculate the maximum theoretical efficiency of a Carnot engine?

The maximum theoretical efficiency of a Carnot engine is calculated using the temperatures of the hot and cold reservoirs. The formula is:

$\eta =1-\frac{T_c}{T_h}$

where $T_c$ is the temperature of the cold reservoir and $T_h$ is the temperature of the hot reservoir. Both temperatures must be in absolute units (Kelvin). This equation shows that the efficiency depends only on the temperatures of the reservoirs and not on the working substance or the specific details of the engine.

What are the four steps of the Carnot cycle?

The Carnot cycle consists of four steps:

1. Isothermal Expansion: The gas expands isothermally at the temperature of the hot reservoir, absorbing heat $Q_h$.
2. Adiabatic Expansion: The gas continues to expand without heat transfer, causing its temperature to drop to that of the cold reservoir.
3. Isothermal Compression: The gas is compressed isothermally at the temperature of the cold reservoir, releasing heat $Q_c$.
4. Adiabatic Compression: The gas is compressed without heat transfer, raising its temperature back to that of the hot reservoir.

These steps form a closed loop on a pressure-volume (PV) diagram, and the area enclosed by the loop represents the work done by the engine.

Why can't any real engine achieve the efficiency of a Carnot engine?

No real engine can achieve the efficiency of a Carnot engine because the Carnot cycle is an idealized process that assumes reversible processes and no friction or other dissipative effects. In reality, all engines experience some degree of irreversibility due to factors such as friction, heat losses, and finite speed of operation. These irreversibilities reduce the actual efficiency of real engines compared to the theoretical maximum efficiency predicted by the Carnot cycle. Additionally, achieving the infinitely slow processes required for reversibility is impractical in real-world applications.

How does the temperature difference between reservoirs affect the efficiency of a Carnot engine?

The efficiency of a Carnot engine is directly influenced by the temperature difference between the hot and cold reservoirs. The greater the temperature difference, the higher the efficiency. This relationship is expressed by the equation:

$\eta =1-\frac{T_c}{T_h}$

where $T_c$ is the temperature of the cold reservoir and $T_h$ is the temperature of the hot reservoir. As the temperature of the hot reservoir increases or the temperature of the cold reservoir decreases, the efficiency of the Carnot engine increases. Conversely, if the temperature difference is small, the efficiency will be lower.