Problem 15 - University Physics with Modern Physics

Question

A Carnot engine has an efficiency of 66% and performs 2.5 \(\times\) 10\(^4\) 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\(^\circ\)C). What is the temperature of its heat source?

Step-by-Step Solution

Answer

(a) Heat extracted is approximately 3.79 × 10⁴ J. (b) Temperature of the heat source is about 862.21 K.

Step 1: Understanding Efficiency

The efficiency of a Carnot engine is given by the formula \( \eta = 1 - \frac{T_c}{T_h} \), where \( T_c \) and \( T_h \) are the temperatures of the cold and hot reservoirs, respectively. Given the efficiency \( \eta = 0.66 \), which means 66%, we'll use this information later to find the temperature of the heat source.

Step 2: Calculating Heat Energy Input

Efficiency \( \eta \) is also defined as the ratio of work done \( W \) to the heat absorbed from the heat source \( Q_h \). The equation is \( \eta = \frac{W}{Q_h} \). Given \( W = 2.5 \times 10^4 \) J and \( \eta = 0.66 \), we solve for \( Q_h \) using \( 0.66 = \frac{2.5 \times 10^4}{Q_h} \). Rearranging gives \( Q_h = \frac{2.5 \times 10^4}{0.66} \approx 3.79 \times 10^4 \) J.

Step 3: Understanding Temperatures in Kelvin

Convert the exhaust temperature from Celsius to Kelvin since thermodynamic calculations require the temperature in Kelvin. The conversion is given by \( T(K) = T(^{\circ}C) + 273.15 \). For a room temperature of 20.0°C, we have \( T_c = 20 + 273.15 = 293.15 \) K.

Step 4: Finding the Temperature of the Heat Source

Using the Carnot efficiency equation \( \eta = 1 - \frac{T_c}{T_h} \), where \( \eta = 0.66 \) and \( T_c = 293.15 \) K, we can find \( T_h \). Solving \( 0.66 = 1 - \frac{293.15}{T_h} \) yields \( \frac{293.15}{T_h} = 0.34 \), and rearranging gives \( T_h = \frac{293.15}{0.34} \approx 862.21 \) K.

Key Concepts

Efficiency of Heat Engines

In the world of thermodynamics, the efficiency of a heat engine is a crucial aspect to understand. A heat engine extracts heat from a source, performs work, and exhausts some of the heat to a sink. The efficiency (\(\eta\)) measures how well an engine converts the absorbed heat (\(Q_h\)) into work (\(W\)). It can be calculated using the formula:
\(\eta = \frac{W}{Q_h}\).
For a Carnot engine, which is a theoretical perfect engine, the efficiency is also given by the expression:
\(\eta = 1 - \frac{T_c}{T_h}\).
Here, \(T_c\) is the temperature of the cold reservoir and \(T_h\) is the temperature of the hot source. All these temperatures must be in Kelvin.
A Carnot engine represents an ideal because, in reality, 100% efficiency is unattainable due to the second law of thermodynamics. This means that no engine can convert all the heat it absorbs into work without losing some to the surroundings.

Thermodynamic Temperature Conversion

When dealing with thermodynamic processes, it's essential to use the Kelvin scale for temperatures. The Kelvin scale begins at absolute zero, the coldest possible temperature, which helps eliminate negative values that can complicate calculations.
To convert Celsius to Kelvin, simply add 273.15 to the Celsius temperature. For example, converting the room temperature of 20.0°C, we do:

• \(T(K) = T(^{\circ}C) + 273.15\)
• \(T(K) = 20 + 273.15 = 293.15\) K

In thermodynamic calculations involving engines, using Kelvin ensures consistency in calculating formulas like the Carnot efficiency. This temperature conversion is important because avoiding negative values allows for smoother and more precise calculations when using scientific equations.

Work and Heat Energy Relationship

In the study of heat engines, understanding how work and heat energy are interconnected is indispensable. The energy absorbed as heat from the heat source (\(Q_h\)) and the energy lost to the cold sink influence the amount of work (\(W\)) the engine can perform.
An important relationship here is: \(W = Q_h - Q_c\), where \(Q_c\) is the heat expelled to the cold sink. This equation demonstrates that the work done by the engine is the difference between the heat absorbed from the source and the heat lost to the sink.
Moreover, we explored the efficiency calculation \(\eta = \frac{W}{Q_h}\), concluding that efficiency helps determine how much of the absorbed heat is converted to useful work. By improving efficiency, an engine can perform more work for the same amount of heat input, but natural limitations prevent reaching the theoretical 100% efficiency.