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Please explain how you would get part b
![### Thermodynamics Quiz
An engine removes 391 J from a reservoir at 302 K and exhausts 341 J to a reservoir at 200 K.
#### 1) (a) What is its efficiency?
Input Field:
```
12.7 % [Submit]
```
*Note:*
You currently have 1 submission for this question. Only 10 submissions are allowed. You can make 9 more submissions for this question.
---
#### 2) (b) How much more work could be done if the engine were reversible?
Input Field:
```
J [Submit]
```
*Note:*
You currently have 3 submissions for this question. Only 10 submissions are allowed. You can make 7 more submissions for this question.
---
### Explanation of the Problem:
When an engine operates between two thermal reservoirs, the efficiency of the engine can be calculated using the input and output energies and temperatures of the reservoirs. For a non-reversible engine, the efficiency can be less than that of the most efficient possible (reversible) engine operating between the same two temperatures.
#### Key Definitions:
- **Efficiency (\(\eta\))**: The ratio of the work output to the heat input.
\[
\eta = \left( \frac{Work\ Output}{Heat\ Input} \right)
\]
- **Reversible Process**: An idealized process that occurs without any increase in entropy, meaning it can theoretically be reversed without any loss of energy.
#### Concepts:
- The actual efficiency is given and it may be calculated as:
\[
\eta = \left( \frac{391\ J - 341\ J}{391\ J} \right) \times 100\%
\]
- For a reversible engine, the efficiency between two reservoirs can be maximized and computed using Carnot Efficiency:
\[
\eta_{Carnot} = 1 - \left( \frac{T_{cold}}{T_{hot}} \right)
\]
Where \(T_{cold}\) and \(T_{hot}\) are the absolute temperatures of the cold and hot reservoirs respectively.
The work done by a reversible engine can be more than the work done by a non-reversible one due to zero entropy change.
By comparing non-reversible and reversible processes, you can determine the potential increase in work output.](https://content.bartleby.com/qna-images/question/e6e52c04-e2c3-4e2d-83a7-f5740c8e0fd5/71e79df0-32d5-411d-a9e2-9d39d1f3db56/de6omdn_processed.jpeg)
Transcribed Image Text:### Thermodynamics Quiz
An engine removes 391 J from a reservoir at 302 K and exhausts 341 J to a reservoir at 200 K.
#### 1) (a) What is its efficiency?
Input Field:
```
12.7 % [Submit]
```
*Note:*
You currently have 1 submission for this question. Only 10 submissions are allowed. You can make 9 more submissions for this question.
---
#### 2) (b) How much more work could be done if the engine were reversible?
Input Field:
```
J [Submit]
```
*Note:*
You currently have 3 submissions for this question. Only 10 submissions are allowed. You can make 7 more submissions for this question.
---
### Explanation of the Problem:
When an engine operates between two thermal reservoirs, the efficiency of the engine can be calculated using the input and output energies and temperatures of the reservoirs. For a non-reversible engine, the efficiency can be less than that of the most efficient possible (reversible) engine operating between the same two temperatures.
#### Key Definitions:
- **Efficiency (\(\eta\))**: The ratio of the work output to the heat input.
\[
\eta = \left( \frac{Work\ Output}{Heat\ Input} \right)
\]
- **Reversible Process**: An idealized process that occurs without any increase in entropy, meaning it can theoretically be reversed without any loss of energy.
#### Concepts:
- The actual efficiency is given and it may be calculated as:
\[
\eta = \left( \frac{391\ J - 341\ J}{391\ J} \right) \times 100\%
\]
- For a reversible engine, the efficiency between two reservoirs can be maximized and computed using Carnot Efficiency:
\[
\eta_{Carnot} = 1 - \left( \frac{T_{cold}}{T_{hot}} \right)
\]
Where \(T_{cold}\) and \(T_{hot}\) are the absolute temperatures of the cold and hot reservoirs respectively.
The work done by a reversible engine can be more than the work done by a non-reversible one due to zero entropy change.
By comparing non-reversible and reversible processes, you can determine the potential increase in work output.
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