A Carnot engine rejects 50% of the heat absorbed from a source to a sink at 27°C. What is the source temperature?
A Carnot engine rejects 50% of the heat absorbed from a source to a sink at 27°C. What is the source temperature?
Elements Of Electromagnetics
7th Edition
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
ChapterMA: Math Assessment
Section: Chapter Questions
Problem 1.1MA
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![### Problem Statement
A Carnot engine rejects 50% of the heat absorbed from a source to a sink at 27°C. What is the source temperature?
### Explanation
The problem describes a scenario involving a Carnot engine, which is a theoretical model of a thermodynamic engine used to explain the principles of thermodynamic cycles. The Carnot engine operates between a heat source and a heat sink, rejecting a certain percentage of heat absorbed from the source to the sink. In this problem, the heat sink is at 27°C.
To solve this, we need to employ the Carnot efficiency formula and the relationship between the temperatures of the source and the sink.
### Key Concepts
1. **Carnot Efficiency**: The efficiency \(\eta\) of a Carnot engine is given by:
\[ \eta = 1 - \frac{T_\text{sink}}{T_\text{source}} \]
2. **Percentage of Heat Rejection**: If 50% of the heat is rejected, then \(\eta = 0.50\).
3. **Temperature Conversion**: Temperatures must be converted to an absolute scale (Kelvin, K) for calculations:
\[ T(K) = T(°C) + 273.15 \]
### Steps to Solve:
1. Convert the sink temperature to Kelvin:
\[ T_\text{sink} = 27 + 273.15 = 300.15\,K \]
2. Use the efficiency formula:
\[ \eta = 1 - \frac{T_\text{sink}}{T_\text{source}} \]
Since the Carnot engine rejects 50% of the heat, the efficiency \(\eta = 0.50\):
\[ 0.50 = 1 - \frac{300.15}{T_\text{source}} \]
3. Solve for \(T_\text{source}\):
\[ \frac{300.15}{T_\text{source}} = 0.50 \]
\[ T_\text{source} = \frac{300.15}{0.50} \]
\[ T_\text{source} = 600.30\,K \]
4. Convert back to Celsius if required:
\[ T_\text{source}(°C) = T_\text{source}(K) - 273.15 \]
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Transcribed Image Text:### Problem Statement
A Carnot engine rejects 50% of the heat absorbed from a source to a sink at 27°C. What is the source temperature?
### Explanation
The problem describes a scenario involving a Carnot engine, which is a theoretical model of a thermodynamic engine used to explain the principles of thermodynamic cycles. The Carnot engine operates between a heat source and a heat sink, rejecting a certain percentage of heat absorbed from the source to the sink. In this problem, the heat sink is at 27°C.
To solve this, we need to employ the Carnot efficiency formula and the relationship between the temperatures of the source and the sink.
### Key Concepts
1. **Carnot Efficiency**: The efficiency \(\eta\) of a Carnot engine is given by:
\[ \eta = 1 - \frac{T_\text{sink}}{T_\text{source}} \]
2. **Percentage of Heat Rejection**: If 50% of the heat is rejected, then \(\eta = 0.50\).
3. **Temperature Conversion**: Temperatures must be converted to an absolute scale (Kelvin, K) for calculations:
\[ T(K) = T(°C) + 273.15 \]
### Steps to Solve:
1. Convert the sink temperature to Kelvin:
\[ T_\text{sink} = 27 + 273.15 = 300.15\,K \]
2. Use the efficiency formula:
\[ \eta = 1 - \frac{T_\text{sink}}{T_\text{source}} \]
Since the Carnot engine rejects 50% of the heat, the efficiency \(\eta = 0.50\):
\[ 0.50 = 1 - \frac{300.15}{T_\text{source}} \]
3. Solve for \(T_\text{source}\):
\[ \frac{300.15}{T_\text{source}} = 0.50 \]
\[ T_\text{source} = \frac{300.15}{0.50} \]
\[ T_\text{source} = 600.30\,K \]
4. Convert back to Celsius if required:
\[ T_\text{source}(°C) = T_\text{source}(K) - 273.15 \]
\[
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