A beat engine work on a Cannot cycle with the heat sink in the temperature of `27^(@)C`. IF the efficiency is 20%, then the temperature (in Kelvin) of the heat source will be

Options

A. 375
B. 300
C. 270
D. 150

Answer

To solve the problem, we need to find the temperature of the heat source (T1) for a Carnot engine given the efficiency and the temperature of the heat sink (T2). Here’s the step-by-step solution:

### Step 1: Convert the temperature of the heat sink to Kelvin
The temperature of the heat sink (T2) is given as 27°C. To convert this to Kelvin, we use the formula:
\[ T(K) = T(°C) + 273 \]
So,
\[ T2 = 27 + 273 = 300 \, K \]

**Hint:** Remember to convert Celsius to Kelvin by adding 273.

### Step 2: Write the formula for efficiency of a Carnot engine
The efficiency (η) of a Carnot engine is given by the formula:
\[ \eta = 1 - \frac{T2}{T1} \]
Where:
- η is the efficiency
- T2 is the temperature of the heat sink
- T1 is the temperature of the heat source

**Hint:** Efficiency is a measure of how well an engine converts heat from the heat source into work.

### Step 3: Substitute the known values into the efficiency formula
We know the efficiency is 20%, which can be expressed as a decimal:
\[ \eta = \frac{20}{100} = 0.2 \]
Now substituting the values into the formula:
\[ 0.2 = 1 - \frac{300}{T1} \]

**Hint:** Make sure to convert percentage efficiency into decimal form before using it in calculations.

### Step 4: Rearrange the equation to solve for T1
Rearranging the equation gives:
\[ \frac{300}{T1} = 1 - 0.2 \]
\[ \frac{300}{T1} = 0.8 \]

Now, cross-multiply to find T1:
\[ 300 = 0.8 \times T1 \]
\[ T1 = \frac{300}{0.8} \]

**Hint:** When rearranging equations, ensure to isolate the variable you are solving for.

### Step 5: Calculate T1
Now, calculating T1:
\[ T1 = \frac{300}{0.8} = 375 \, K \]

**Hint:** When dividing, ensure to perform the arithmetic carefully to avoid errors.

### Conclusion
The temperature of the heat source (T1) is 375 K.

**Final Answer:** T1 = 375 K