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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### 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} =
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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