The efficiency of an automobile engine increases by 5.0%. For an input heat of 1300 J, how much more work does the engine produce?

Options

A. 26 J
B. 65 J
C. 41 J
D. 88 J

Answer

To solve the problem of how much more work an automobile engine produces when its efficiency increases by 5% with an input heat of 1300 J, we can follow these steps:

### Step 1: Understand the relationship between efficiency, work, and heat input
The efficiency (η) of a heat engine is defined as the ratio of the work done (W) to the heat input (Q_in):
\[ \eta = \frac{W}{Q_{in}} \]

### Step 2: Express work in terms of efficiency and heat input
From the definition of efficiency, we can rearrange the equation to express work:
\[ W = \eta \cdot Q_{in} \]

### Step 3: Determine the initial efficiency
Let’s denote the initial efficiency as η_initial. If the efficiency increases by 5%, the new efficiency (η_new) can be expressed as:
\[ \eta_{new} = \eta_{initial} + 0.05 \]

### Step 4: Calculate the initial work done
Using the initial efficiency, the work done initially (W_initial) can be calculated as:
\[ W_{initial} = \eta_{initial} \cdot Q_{in} \]

### Step 5: Calculate the new work done
Using the new efficiency, the work done after the increase in efficiency (W_new) can be calculated as:
\[ W_{new} = \eta_{new} \cdot Q_{in} \]

### Step 6: Find the change in work
The change in work (ΔW) can be found by subtracting the initial work from the new work:
\[ \Delta W = W_{new} - W_{initial} \]

### Step 7: Substitute the values
Since we know that:
\[ W_{new} = (\eta_{initial} + 0.05) \cdot Q_{in} \]
\[ W_{initial} = \eta_{initial} \cdot Q_{in} \]
Thus,
\[ \Delta W = [(\eta_{initial} + 0.05) \cdot Q_{in}] - [\eta_{initial} \cdot Q_{in}] \]
\[ \Delta W = 0.05 \cdot Q_{in} \]

### Step 8: Plug in the input heat
Given that \(Q_{in} = 1300 \, J\):
\[ \Delta W = 0.05 \cdot 1300 \]
\[ \Delta W = 65 \, J \]

### Conclusion
The increase in work produced by the engine due to the increase in efficiency is **65 J**.

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