Problem 15 - University Physics with Modern Physics

Question

20.15. A Carnot engine has an efficiency of 59\% and performs \(2.5 \times 10^{4} \mathrm{J}\) of work in each cycle. (a) How much heat does the engine extract from its heat source in each cycle? (b) Suppose the engine exhausts heat at room temperature ( \(20.0^{\circ} \mathrm{C} )\) . What is the temperature of its heat source?

Step-by-Step Solution

Answer

(a) Heat extracted, \(Q_H = 4.24 \times 10^4 \mathrm{J}\). (b) Heat source temperature, \(T_H \approx 715 \mathrm{K}\).

Step 1: Identify the Efficiency Formula

The efficiency (\(\eta\)) of a Carnot engine is given by the formula \(\eta = \frac{W}{Q_H}\), where \(W\) is the work done and \(Q_H\) is the heat extracted from the hot reservoir. We are told that \(\eta = 59\% = 0.59\) and \(W = 2.5 \times 10^4\, \mathrm{J}\).

Step 2: Calculate Heat Extracted

Rearrange the efficiency formula to solve for \(Q_H\). \(Q_H = \frac{W}{\eta}\). Substitute the known values: \(Q_H = \frac{2.5 \times 10^4}{0.59}\). Calculate \(Q_H\).

Step 3: Efficiency and Temperature Relation

The efficiency of a Carnot engine is also given by \(\eta = 1 - \frac{T_C}{T_H}\), where \(T_C\) is the temperature of the cold reservoir and \(T_H\) is the temperature of the hot reservoir, both in Kelvin. Room temperature \(T_C\) is given as \(20.0^{\circ} \mathrm{C}\), which is \(273.15 + 20.0 = 293.15 \mathrm{K}\).

Step 4: Determine Temperature of Heat Source

Rearrange the efficiency-temperature relation to solve for \(T_H\): \(T_H = \frac{T_C}{1 - \eta}\). Substitute the values: \(T_H = \frac{293.15}{1 - 0.59}\). Calculate \(T_H\).

Key Concepts

Efficiency of a Carnot Engine

Efficiency is a measure of how well an engine converts heat energy into work. In the specific case of the Carnot engine, its efficiency is defined as the ratio of the work output to the heat input from the hot reservoir. The formula used is: \[ \eta = \frac{W}{Q_H} \]where \( \eta \) is the efficiency, \( W \) is the work done, and \( Q_H \) is the heat drawn from the heat source. In this exercise, the Carnot engine's efficiency is given as 59%, meaning it can convert 59% of the heat energy from the heat source into useful work.
This leaves 41% of the energy that must be expelled as waste heat. Understanding this efficiency allows us to better understand the constraints and limitations of heat engines.

Heat Source in Thermodynamics

The heat source is a critical component of a Carnot engine. It provides the thermal energy required for the engine to perform work. In this exercise, calculating the heat extracted from the heat source gives additional insight into how much energy is required for the engine operation.Using the formula: \[ Q_H = \frac{W}{\eta} \]we can determine the energy input from the heat source given the efficiency and work output. The heat source's quality (e.g., temperature) significantly influences the engine's capability to convert heat into work efficiently.

Role of Temperature in Engine Efficiency

Temperature plays a vital role in determining the efficiency of a Carnot engine. The efficiency is dependent on the temperatures of both the hot and cold reservoirs. The relationship is given by the formula:\[ \eta = 1 - \frac{T_C}{T_H} \]where \( T_C \) is the cold reservoir temperature and \( T_H \) is the heat source's temperature. These temperatures are measured in Kelvin to facilitate accurate calculations. In this scenario, the temperature of the cold reservoir is room temperature (20°C), which is equivalent to 293.15 K.
This formula helps us understand that the higher the heat source temperature compared to the cold reservoir, the higher the efficiency. Engineers often strive to increase the heat source temperature to achieve greater efficiency.

Understanding Thermodynamics and the Carnot Engine

Thermodynamics is the science of energy transfer, and when applied to engines, it primarily deals with how heat energy is converted to work and vice versa. The Carnot engine is an idealized model that provides the maximum possible efficiency for a heat engine operating between two thermal reservoirs. This is rooted in two main laws:

• The First Law of Thermodynamics, which is essentially a version of the law of energy conservation adapted for thermodynamic systems.
• The Second Law of Thermodynamics, which introduces the concept of irreversibility and provides the foundation for the principle that no engine can be more efficient than a Carnot engine operating between the same two thermal reservoirs.

Understanding these principles helps explain why real-world engines can never quite reach the efficiency of a Carnot engine, as they are always subject to inefficiencies due to factors like friction and heat losses.