Problem 22 - College Physics

Question

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 \(\left(20.0^{\circ} \mathrm{C}\right) .\) What is the temperature of its heat source?

Step-by-Step Solution

Answer

(a) 4.24 × 10⁴ J; (b) 714.02 K

Step 1: Understand Efficiency Formula

The efficiency \( \eta \) of a Carnot engine 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 hot reservoir temperature. Also, efficiency can be given by \( \eta = \frac{W}{Q_h} \), where \( W \) is the work done, and \( Q_h \) is the heat extracted from the hot reservoir.

Step 2: Calculate Heat Extracted From Heat Source

Given the work \( W = 2.5 \times 10^4 \text{ J} \) and efficiency \( \eta = 0.59 \), use \( \eta = \frac{W}{Q_h} \). Rearranging, we find \( Q_h = \frac{W}{\eta} \). Substituting the known values gives \( Q_h = \frac{2.5 \times 10^4}{0.59} \approx 4.24 \times 10^4 \text{ J} \).

Step 3: Convert Room Temperature to Kelvin

The room temperature is given as \( 20.0^{\circ} \text{C} \). Convert this to Kelvin using the formula \( T(K) = T(^{\circ}\text{C}) + 273.15 \). Therefore, \( T_c = 20.0 + 273.15 = 293.15 \text{ K} \).

Step 4: Determine the Temperature of the Heat Source

We know \( \eta = 1 - \frac{T_c}{T_h} \). Rearranging this gives \( T_h = \frac{T_c}{1 - \eta} \). Substituting \( T_c = 293.15 \text{ K} \) and \( \eta = 0.59 \), we calculate \( T_h = \frac{293.15}{1 - 0.59} \approx 714.02 \text{ K} \).

Key Concepts

Thermodynamics

Thermodynamics is a branch of physics that deals with the relationships between heat, work, temperature, and energy. It explains how energy is transformed from one form to another and how it affects matter. The core principle revolves around the laws of thermodynamics.

• **First Law**: Energy cannot be created or destroyed, only transformed.
• **Second Law**: Heat energy always flows from a hotter object to a cooler one unless work is done on the system.
• **Third Law**: As a system approaches absolute zero, the entropy, or disorder, approaches a minimum value.

These laws help us understand processes involving energy exchanges, like how engines convert heat to work and influence the efficiency of appliances and industrial processes.

Carnot Cycle

The Carnot cycle is a theoretical model that defines the maximum possible efficiency an engine can achieve by operating between two heat reservoirs.

• A Carnot cycle consists of four reversible stages: two isothermal (constant temperature) processes and two adiabatic (no heat loss or gain) processes.
• During isothermal expansion, the system does work while absorbing heat from the hot reservoir.
• In adiabatic expansion, the system continues to do work without transferring heat, causing its temperature to drop until it matches the cold reservoir.
• Isothermal compression follows, where the system releases heat to the cold reservoir.
• Finally, adiabatic compression increases the system's temperature back to the initial state, completing the cycle.

These cycles show the limitations of heat engines, demonstrating the highest efficiency that can be theoretically achieved.

Temperature Conversion

Temperature conversion is essential when calculating efficiencies and understanding heat engines. In thermodynamics, temperature is often measured in Kelvin, which is the absolute temperature scale.
To convert Celsius to Kelvin, the formula used is:\[T(K) = T(^{\circ}C) + 273.15\]This conversion is straightforward but crucial, especially for applying the Carnot efficiency formula, \[\eta = 1 - \frac{T_c}{T_h}\]where temperatures must be in Kelvin. Using the correct units ensures the mathematical relationships reflect realistic physical behaviors and lead to accurate calculations.

Heat Energy

Heat energy plays a central role in how engines and many natural processes operate. It represents energy transfer between systems due to temperature differences.
In a Carnot engine, heat is absorbed from a hot reservoir into the engine and partly converted into work (useful mechanical energy), while the rest is exhausted to a cold reservoir.
Calculating heat transfer involves understanding the efficiency of the engine. The efficiency equation \(\eta = \frac{W}{Q_h}\), where \(W\) is the work done, and \(Q_h\) is the heat extracted from the hot source, helps determine how much energy is wasted or used effectively.
Knowing these relationships aids in optimizing engine design and improving energy efficiency in various applications.