5. (15.22) A heat engine's high temperature TH could be ambient temperature, because liquid nitrogen at 77 K could be Ti and is cheap. What would be the efficiency of a Camot engine that made use of heat transferred from air at room temperature (293 K) to the liquid nitrogen "fuel"?
5. (15.22) A heat engine's high temperature TH could be ambient temperature, because liquid nitrogen at 77 K could be Ti and is cheap. What would be the efficiency of a Camot engine that made use of heat transferred from air at room temperature (293 K) to the liquid nitrogen "fuel"?
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Author:Raymond A. Serway, Chris Vuille
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Chapter1: Units, Trigonometry. And Vectors
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![### Thermodynamics Problems for Educational Purposes
#### Problem 5
**(15.22)** A heat engine's high temperature \( T_H \) could be ambient temperature because liquid nitrogen at 77 K could be \( T_L \) and is cheap. What would be the efficiency of a Carnot engine that made use of heat transferred from air at room temperature (293 K) to the liquid nitrogen "fuel"?
#### Problem 6
**(15.32)** If an ideal refrigerator keeps its contents at 2.5°C when the house temperature is 22°C, what is its COP (Coefficient of Performance)?
#### Problem 7
**(15.39)** What is the change in entropy of 320 g of steam at 100°C when it is condensed to water at 100°C?
### Explanation of Concepts
#### Carnot Engine Efficiency
The efficiency \(\eta\) of a Carnot engine is given by the formula:
\[
\eta = 1 - \frac{T_L}{T_H}
\]
where \( T_H \) is the high temperature reservoir and \( T_L \) is the low temperature reservoir. Both temperatures must be in Kelvin.
#### Coefficient of Performance (COP) of a Refrigerator
The Coefficient of Performance (COP) for a refrigerator is given by:
\[
COP = \frac{T_L}{T_H - T_L}
\]
where \( T_L \) is the temperature inside the refrigerator and \( T_H \) is the ambient temperature. Both temperatures must be converted to Kelvin.
#### Change in Entropy
The change in entropy \(\Delta S\) when steam condenses to water is calculated by:
\[
\Delta S = \frac{Q}{T}
\]
where \( Q \) is the heat released during condensation, and \( T \) is the temperature at which the process occurs (in Kelvin).
### Example Calculations
#### Problem 5
Given:
- \( T_H = 293 \) K (room temperature)
- \( T_L = 77 \) K (temperature of liquid nitrogen)
The Carnot efficiency:
\[
\eta = 1 - \frac{77}{293}
\]
#### Problem 6
Given:
- \( T_L = 2.5^\circ \text{C} = 275.65 \) K
- \( T_H = 22^\circ \text{C} =](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F529e4fe6-e8c4-4722-88aa-75bc8f350b68%2Fc0acf8ad-a009-41b3-984a-28f262206df1%2Fcdh7vbl_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Thermodynamics Problems for Educational Purposes
#### Problem 5
**(15.22)** A heat engine's high temperature \( T_H \) could be ambient temperature because liquid nitrogen at 77 K could be \( T_L \) and is cheap. What would be the efficiency of a Carnot engine that made use of heat transferred from air at room temperature (293 K) to the liquid nitrogen "fuel"?
#### Problem 6
**(15.32)** If an ideal refrigerator keeps its contents at 2.5°C when the house temperature is 22°C, what is its COP (Coefficient of Performance)?
#### Problem 7
**(15.39)** What is the change in entropy of 320 g of steam at 100°C when it is condensed to water at 100°C?
### Explanation of Concepts
#### Carnot Engine Efficiency
The efficiency \(\eta\) of a Carnot engine is given by the formula:
\[
\eta = 1 - \frac{T_L}{T_H}
\]
where \( T_H \) is the high temperature reservoir and \( T_L \) is the low temperature reservoir. Both temperatures must be in Kelvin.
#### Coefficient of Performance (COP) of a Refrigerator
The Coefficient of Performance (COP) for a refrigerator is given by:
\[
COP = \frac{T_L}{T_H - T_L}
\]
where \( T_L \) is the temperature inside the refrigerator and \( T_H \) is the ambient temperature. Both temperatures must be converted to Kelvin.
#### Change in Entropy
The change in entropy \(\Delta S\) when steam condenses to water is calculated by:
\[
\Delta S = \frac{Q}{T}
\]
where \( Q \) is the heat released during condensation, and \( T \) is the temperature at which the process occurs (in Kelvin).
### Example Calculations
#### Problem 5
Given:
- \( T_H = 293 \) K (room temperature)
- \( T_L = 77 \) K (temperature of liquid nitrogen)
The Carnot efficiency:
\[
\eta = 1 - \frac{77}{293}
\]
#### Problem 6
Given:
- \( T_L = 2.5^\circ \text{C} = 275.65 \) K
- \( T_H = 22^\circ \text{C} =
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