An engine that operates at half its theoretical (Carnot) efficiency, operates between `545^@C` and `310^@C` while producing work at the rate of 1000 kW. How much heat is discharged per hour?

Answer

To solve the problem step by step, we will follow the principles of thermodynamics and the equations related to the Carnot efficiency.

### Step 1: Convert temperatures from Celsius to Kelvin
The temperatures given are:
- \( T_H = 545^\circ C \)
- \( T_C = 310^\circ C \)

To convert these temperatures to Kelvin, we use the formula:
\[ T(K) = T(°C) + 273.15 \]

Calculating:
\[ T_H = 545 + 273.15 = 818.15 \, K \]
\[ T_C = 310 + 273.15 = 583.15 \, K \]

### Step 2: Calculate the theoretical (Carnot) efficiency
The efficiency of a Carnot engine is given by the formula:
\[ \eta = 1 - \frac{T_C}{T_H} \]

Substituting the values:
\[ \eta = 1 - \frac{583.15}{818.15} \]

Calculating:
\[ \eta \approx 1 - 0.712 = 0.288 \, \text{or} \, 28.8\% \]

### Step 3: Find the actual efficiency of the engine
The engine operates at half of its theoretical efficiency:
\[ \eta_{actual} = \frac{1}{2} \times \eta_{theoretical} = \frac{1}{2} \times 0.288 = 0.144 \, \text{or} \, 14.4\% \]

### Step 4: Calculate the heat input (Q1)
The work done by the engine is given as \( W = 1000 \, kW = 1000 \, kJ/s \).

Using the efficiency formula:
\[ \eta = \frac{W}{Q_1} \]
Rearranging gives:
\[ Q_1 = \frac{W}{\eta} \]

Substituting the values:
\[ Q_1 = \frac{1000 \, kJ/s}{0.144} \approx 6944.44 \, kJ/s \]

### Step 5: Calculate the heat discharged (Q2)
The heat discharged can be calculated using:
\[ Q_2 = Q_1 - W \]

Substituting the values:
\[ Q_2 = 6944.44 \, kJ/s - 1000 \, kJ/s = 5944.44 \, kJ/s \]

### Step 6: Convert heat discharged to per hour
To find the heat discharged per hour, we multiply by the number of seconds in an hour (3600 seconds):
\[ Q_2 \, (\text{per hour}) = 5944.44 \, kJ/s \times 3600 \, s/h \]

Calculating:
\[ Q_2 \, (\text{per hour}) = 5944.44 \times 3600 \approx 21,422,000 \, kJ/h \]

### Final Answer
The heat discharged per hour is approximately:
\[ Q_2 \approx 21.42 \times 10^6 \, kJ/h \]

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