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 find the maximum work that can be obtained from the engine, we can use the concept of efficiency of a heat engine operating between two reservoirs. The efficiency (η) of a Carnot engine is given by the formula:

\[ \eta = 1 - \frac{T_c}{T_h} \]

where:
- \(T_c\) is the temperature of the cold reservoir (in Kelvin),
- \(T_h\) is the temperature of the hot reservoir (in Kelvin).

### Step 1: Identify the temperatures
From the problem:
- Hot reservoir temperature, \(T_h = 650 \, K\)
- Cold reservoir temperature, \(T_c = 150 \, K\)

### Step 2: Calculate the efficiency
Using the efficiency formula:

\[ \eta = 1 - \frac{T_c}{T_h} = 1 - \frac{150}{650} \]

### Step 3: Simplify the fraction
To simplify \(\frac{150}{650}\):

\[ \frac{150}{650} = \frac{15}{65} = \frac{3}{13} \]

### Step 4: Substitute back into the efficiency equation
Now, substituting back into the efficiency equation:

\[ \eta = 1 - \frac{3}{13} = \frac{13 - 3}{13} = \frac{10}{13} \]

### Step 5: Calculate the maximum work
The maximum work (W) that can be obtained from the engine is given by:

\[ W = \eta \times Q_h \]

where \(Q_h\) is the heat supplied to the engine. Given \(Q_h = 1200 \, J\):

\[ W = \frac{10}{13} \times 1200 \]

### Step 6: Perform the multiplication
Calculating the maximum work:

\[ W = \frac{12000}{13} \]

### Step 7: Calculate the final value
Now, performing the division:

\[ W \approx 923.08 \, J \]

Thus, the maximum work that can be obtained from the engine is approximately:

\[ W \approx 923.08 \, J \]

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

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