In Carnot engine, efficiency is 40% at hot reservoir temperature T. For efficiency 50% , what will be the temperature of hot reservoir?

Options

A. `(2T)/5`
B. `6T`
C. `(6T)/5`
D. `T/5`

Answer

To solve the problem step by step, we will use the formula for the efficiency of a Carnot engine, which is given by:

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

where \( T_1 \) is the temperature of the hot reservoir and \( T_2 \) is the temperature of the cold reservoir.

### Step 1: Set up the equation for the first efficiency (40%)
Given that the efficiency is 40%, we can express this as:

\[ \text{Efficiency} = 0.4 = 1 - \frac{T_2}{T} \]

Here, we assume the temperature of the hot reservoir \( T_1 = T \).

### Step 2: Rearrange the equation to find \( T_2 \)
Rearranging the equation gives:

\[ 0.4 = 1 - \frac{T_2}{T} \]

Subtracting 1 from both sides:

\[ 0.4 - 1 = -\frac{T_2}{T} \]

This simplifies to:

\[ -0.6 = -\frac{T_2}{T} \]

Multiplying both sides by -1:

\[ 0.6 = \frac{T_2}{T} \]

Now, multiplying both sides by \( T \):

\[ T_2 = 0.6T \]

### Step 3: Set up the equation for the second efficiency (50%)
Now, we need to find the temperature of the hot reservoir when the efficiency is 50%. We can express this as:

\[ \text{Efficiency} = 0.5 = 1 - \frac{T_2}{T_1} \]

### Step 4: Substitute \( T_2 \) from Step 2 into the new equation
We already found \( T_2 = 0.6T \). Substituting this into the equation gives:

\[ 0.5 = 1 - \frac{0.6T}{T_1} \]

### Step 5: Rearrange the equation to find \( T_1 \)
Rearranging the equation gives:

\[ 0.5 = 1 - \frac{0.6T}{T_1} \]

Subtracting 1 from both sides:

\[ 0.5 - 1 = -\frac{0.6T}{T_1} \]

This simplifies to:

\[ -0.5 = -\frac{0.6T}{T_1} \]

Multiplying both sides by -1:

\[ 0.5 = \frac{0.6T}{T_1} \]

Now, cross-multiplying gives:

\[ 0.5T_1 = 0.6T \]

### Step 6: Solve for \( T_1 \)
Dividing both sides by 0.5:

\[ T_1 = \frac{0.6T}{0.5} = 1.2T \]

### Conclusion
Thus, the temperature of the hot reservoir when the efficiency is 50% is:

\[ T_1 = 1.2T \]