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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? 12.7 % Submit You currently have 1 submissions for this question. Only 10 submission 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? J Submit You currently have 3 submissions for this question. Only 10 submission are allowed. You can make 7 more submissions for this question.

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? 12.7 % Submit You currently have 1 submissions for this question. Only 10 submission 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? J Submit You currently have 3 submissions for this question. Only 10 submission are allowed. You can make 7 more submissions for this question.

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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.
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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