A heat engine released 1020 J of heat and performed 900 J of useful work. A cannot engine operating between the same temperature ranges as this engine is 30% more efficient. If temperature of source is 700 K, temperature of sink is

Options

A. 390 K
B. 280 K
C. 460 K
D. 170 K

Answer

To solve the problem, we will follow these steps:

### Step 1: Determine the efficiency of the given heat engine
The efficiency (η) of a heat engine is defined as the ratio of the work done (W) to the heat input (Q_in).

Given:
- Work done (W) = 900 J
- Heat released (Q_out) = 1020 J

First, we need to find the heat input (Q_in):
\[ Q_{in} = W + Q_{out} = 900 J + 1020 J = 1920 J \]

Now, we can calculate the efficiency:
\[ \eta = \frac{W}{Q_{in}} = \frac{900 J}{1920 J} \]

Calculating this gives:
\[ \eta \approx 0.46875 \text{ or } 46.875\% \]

### Step 2: Calculate the efficiency of the Carnot engine
The problem states that the Carnot engine is 30% more efficient than the given heat engine. Therefore, we can express the efficiency of the Carnot engine (η_C) as:
\[ \eta_C = \eta + 0.30 \cdot \eta = \eta \cdot (1 + 0.30) = 0.46875 \cdot 1.30 \]

Calculating this gives:
\[ \eta_C \approx 0.609375 \text{ or } 60.9375\% \]

### Step 3: Relate the Carnot efficiency to the temperatures
The efficiency of a Carnot engine operating between two temperatures is given by:
\[ \eta_C = 1 - \frac{T_{sink}}{T_{source}} \]

Given that the temperature of the source (T_source) is 700 K, we can substitute:
\[ 0.609375 = 1 - \frac{T_{sink}}{700} \]

### Step 4: Solve for the temperature of the sink (T_sink)
Rearranging the equation:
\[ \frac{T_{sink}}{700} = 1 - 0.609375 \]
\[ \frac{T_{sink}}{700} = 0.390625 \]

Now, multiplying both sides by 700:
\[ T_{sink} = 0.390625 \times 700 \]
\[ T_{sink} \approx 273.4375 \text{ K} \]

### Conclusion
The temperature of the sink is approximately 273.43 K.