A heat of 1200 J is supplied to an engine from a hot reservoir maintained at a temperature of 650 K. A 150 K reservoir is used as the cold reservoir. What is the maximum work (in J) that can be obtained from the engine?

Explanation

To solve the problem of finding the maximum work that can be obtained from the engine, we can follow these steps:

### Step 1: Identify the temperatures of the reservoirs
- The hot reservoir temperature \( T_1 = 650 \, \text{K} \)
- The cold reservoir temperature \( T_2 = 150 \, \text{K} \)

### Step 2: Calculate the efficiency of the heat engine
The efficiency \( \eta \) of a Carnot engine is given by the formula:
\[ \eta = 1 - \frac{T_2}{T_1} \]
Substituting the values:
\[ \eta = 1 - \frac{150}{650} \]

### Step 3: Simplify the efficiency calculation
Calculating the fraction:
\[ \frac{150}{650} = 0.230769 \]
Thus,
\[ \eta = 1 - 0.230769 = 0.769231 \]
Expressing this as a percentage:
\[ \eta \approx 76.9\% \]

### Step 4: Calculate the maximum work done by the engine
The maximum work \( W \) done by the engine can be calculated using the formula:
\[ W = \eta \times Q_1 \]
where \( Q_1 \) is the heat supplied to the engine. Given \( Q_1 = 1200 \, \text{J} \):
\[ W = 0.769231 \times 1200 \]

### Step 5: Perform the multiplication
Calculating the work:
\[ W \approx 923.076 \, \text{J} \]
Rounding to the nearest whole number:
\[ W \approx 923 \, \text{J} \]

### Final Answer
The maximum work that can be obtained from the engine is approximately **923 J**.

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