A carnot engine has efficiency of 50% when its sink is at `227^(@)C`. What is the temperature of source

Options

A. `727^(@)C`
B. `454^(@)C`
C. `113.5^(@)C`
D. `500^(@)C`.

Answer

To find the temperature of the source (T_S) for a Carnot engine with an efficiency of 50% and a sink temperature (T_L) of 227°C, we can follow these steps:

### Step 1: Understand the Efficiency of a Carnot Engine
The efficiency (η) of a Carnot engine is given by the formula:
\[ \eta = 1 - \frac{T_L}{T_S} \]
where:
- \(T_L\) is the absolute temperature of the sink in Kelvin.
- \(T_S\) is the absolute temperature of the source in Kelvin.

### Step 2: Convert Sink Temperature to Kelvin
The given sink temperature is 227°C. We need to convert this to Kelvin using the formula:
\[ T(K) = T(°C) + 273.15 \]
So,
\[ T_L = 227 + 273.15 = 500.15 \, K \]

### Step 3: Substitute the Known Values into the Efficiency Formula
We know the efficiency (η) is 50%, which can be expressed as:
\[ \eta = 0.5 \]
Now, substituting the values into the efficiency formula:
\[ 0.5 = 1 - \frac{500.15}{T_S} \]

### Step 4: Rearrange the Equation to Solve for T_S
Rearranging the equation gives:
\[ \frac{500.15}{T_S} = 1 - 0.5 \]
\[ \frac{500.15}{T_S} = 0.5 \]
Now, cross-multiplying to solve for \(T_S\):
\[ 500.15 = 0.5 \times T_S \]
\[ T_S = \frac{500.15}{0.5} = 1000.3 \, K \]

### Step 5: Convert Temperature Back to Celsius (if needed)
If we need to express the temperature in Celsius:
\[ T_S(°C) = T_S(K) - 273.15 = 1000.3 - 273.15 = 727.15°C \]

### Final Answer
The temperature of the source \(T_S\) is approximately 1000.3 K or 727.15°C.

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