Consider a carnot cycle operating between source tempeature 750 K and sink temperature 350 K producing 1.25 J KJ of mechanical work per cycle, the heat transferred to the engine by the reservers

Options

A. 1.34 KJ
B. 2.34 KJ
C. 3.34 KJ
D. 4.34 KJ

Explanation

To solve the problem, we will use the concept of the Carnot cycle and the formula for efficiency. The efficiency of a Carnot engine is given by:

\[ \text{Efficiency} = 1 - \frac{T_2}{T_1} \]

Where:
- \(T_1\) is the temperature of the hot reservoir (source).
- \(T_2\) is the temperature of the cold reservoir (sink).

### Step 1: Identify the given values
- Hot reservoir temperature, \(T_1 = 750 \, K\)
- Cold reservoir temperature, \(T_2 = 350 \, K\)
- Work done per cycle, \(W = 1.25 \, KJ = 1250 \, J\)

### Step 2: Calculate the efficiency of the Carnot engine
Using the efficiency formula:

\[ \text{Efficiency} = 1 - \frac{T_2}{T_1} = 1 - \frac{350}{750} \]

Calculating the fraction:

\[ \frac{350}{750} = \frac{7}{15} \approx 0.4667 \]

Now substituting back into the efficiency formula:

\[ \text{Efficiency} = 1 - 0.4667 \approx 0.5333 \]

### Step 3: Relate efficiency to work done and heat absorbed
The efficiency can also be expressed in terms of work done and heat absorbed from the hot reservoir \(Q_1\):

\[ \text{Efficiency} = \frac{W}{Q_1} \]

Setting the two expressions for efficiency equal to each other:

\[ 0.5333 = \frac{1250}{Q_1} \]

### Step 4: Solve for \(Q_1\)
Rearranging the equation to find \(Q_1\):

\[ Q_1 = \frac{1250}{0.5333} \]

Calculating \(Q_1\):

\[ Q_1 \approx 2343.75 \, J \]

### Step 5: Convert \(Q_1\) to kilojoules
To express \(Q_1\) in kilojoules:

\[ Q_1 \approx 2.34375 \, KJ \approx 2.34 \, KJ \]

### Final Answer
The heat transferred to the engine by the reservoir is approximately \(2.34 \, KJ\).

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