An ideal Carnot's engine whose efficiency 40% receives heat of 500K. If the efficiency is to be 50% then the temperature of sink will be

Options

A. 600K
B. 800K
C. 1000K
D. 250K

Explanation

To solve the problem, we need to find the temperature of the sink (T2) when the efficiency of the Carnot engine is increased to 50%. We will use the formula for the efficiency of a Carnot engine, which is given by:

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

Where:
- \( T_1 \) is the temperature of the heat source (in Kelvin).
- \( T_2 \) is the temperature of the heat sink (in Kelvin).

**Step 1: Identify the given values.**
- Efficiency (η) = 50% = 0.5
- Temperature of the heat source (T1) = 500 K

**Step 2: Substitute the values into the efficiency formula.**
We can rearrange the efficiency formula to solve for \( T_2 \):

\[ η = 1 - \frac{T_2}{T_1} \]

Substituting the known values:

\[ 0.5 = 1 - \frac{T_2}{500} \]

**Step 3: Rearrange the equation to isolate \( T_2 \).**
Subtract 1 from both sides:

\[ 0.5 - 1 = -\frac{T_2}{500} \]

This simplifies to:

\[ -0.5 = -\frac{T_2}{500} \]

Now, multiply both sides by -1:

\[ 0.5 = \frac{T_2}{500} \]

**Step 4: Solve for \( T_2 \).**
Multiply both sides by 500:

\[ T_2 = 0.5 \times 500 \]

Calculating this gives:

\[ T_2 = 250 \, \text{K} \]

**Final Answer:**
The temperature of the sink (T2) is 250 K.

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