Question

Engine efficiencies. Consider a Carnot engine that runs at \(T_{h}=380 \mathrm{~K}\). (a) Compute the efficiency if \(T_{c}=0^{\circ} \mathrm{C}=273 \mathrm{~K}\). (b) Compute the efficiency if \(T_{c}=50^{\circ} \mathrm{C}=323 \mathrm{~K}\).

Short answer

Part (a) \(28.2\text{\%}\), Part (b) \(15.0\text{\%}\).

Step-by-step solution

Understanding Carnot Efficiency

The efficiency \(\text{Eff}\) of a Carnot engine is given by the formula: \[ \text{Eff} = 1 - \frac{T_c}{T_h} \] where \(T_c\) is the cold reservoir temperature and \(T_h\) is the hot reservoir temperature. Make sure temperatures are in Kelvin.

Plugging in Values for Part (a)

Given \(T_h = 380 \text{ K}\) and \(T_c = 273 \text{ K}\), substitute these values into the Carnot efficiency formula: \[ \text{Eff}_{a} = 1 - \frac{273}{380} \]

Simplifying for Part (a)

Calculate the value inside the fraction first: \[ \frac{273}{380} \approx 0.718 \] Subtract this result from 1 to find the efficiency: \[ \text{Eff}_{a} = 1 - 0.718 = 0.282 \text{ or } 28.2\text{\%} \]

Plugging in Values for Part (b)

Given \(T_h = 380 \text{ K}\) and \(T_c = 323 \text{ K}\), substitute these values into the Carnot efficiency formula: \[ \text{Eff}_{b} = 1 - \frac{323}{380} \]

Simplifying for Part (b)

Calculate the value inside the fraction first: \[ \frac{323}{380} \approx 0.850 \] Subtract this result from 1 to find the efficiency: \[ \text{Eff}_{b} = 1 - 0.850 = 0.150 \text{ or } 15.0\text{\%} \]

Key concepts

thermodynamics

Thermodynamics is a branch of physics that deals with heat, work, and energy. It governs how energy is transferred between systems and how it affects matter. Key principles include the laws of thermodynamics:

The First Law states that energy cannot be created or destroyed, only transformed.

The Second Law states that entropy, a measure of randomness or disorder, always increases in an isolated system.

These principles help us understand how engines, including Carnot engines, operate by converting heat into work efficiently.

Carnot cycle

The Carnot cycle is a theoretical cycle that provides the maximum possible efficiency for any heat engine. It is composed of four reversible processes:

1. Isothermal Expansion: The gas expands while absorbing heat from a hot reservoir at a constant temperature (\(T_h\)).

2. Adiabatic Expansion: The gas continues to expand without exchanging heat, causing its temperature to drop.

3. Isothermal Compression: The gas is compressed while releasing heat to a cold reservoir at a constant temperature (\(T_c\)).

4. Adiabatic Compression: The gas is compressed without heat exchange, causing its temperature to rise back to \(T_h\).

By following this cycle, the Carnot engine achieves maximum efficiency determined by the temperatures of the hot and cold reservoirs.

heat engines

Heat engines are devices that convert thermal energy (heat) into mechanical work. They operate between two reservoirs of different temperatures. Key components include:

1. The hot reservoir: Supplies heat energy.

2. The working substance (usually a gas): Undergoes expansion and compression.

3. The cold reservoir: Absorbs released heat.

The performance of heat engines is evaluated using efficiency, which tells us how well the engine converts input heat into work. The formula for Carnot Efficiency is: \[\text{Eff} = 1 - \frac{T_c}{T_h}\]. This indicates that no engine can be more efficient than a Carnot engine cycling between the same two temperatures.

temperature conversion

In thermodynamics, temperature is often measured in Kelvin (K) for calculations. Converting from Celsius to Kelvin is straightforward:

• Degrees Celsius (°C) to Kelvin (K): Add 273.15. For example, 0°C = 273.15 K.
• Degrees Fahrenheit (°F) to Kelvin (K): Convert Fahrenheit to Celsius first, then use the previous step. \( K = \frac{5}{9}(F - 32) + 273.15 \).

For Carnot efficiency calculations, always ensure temperatures are in Kelvin. This standardization helps avoid errors and ensures accurate results. As seen in the exercise, using Kelvin ensures that the formula \(\text{Eff} = 1 - \frac{T_c}{T_h}\) gives the correct efficiency.