A Carnot engine has an efficiency of 0.40. The Kelvin temperature of its hot reservoir is quadrupled, and the Kelvin temperature of its cold reservoir is doubled. What is the efficiency that results from these changes?

Options

A. `0.30`
B. `0.60`
C. `0.50`
D. `0.70`

Answer

To solve the problem, we need to determine the new efficiency of a Carnot engine after changes in the temperatures of its hot and cold reservoirs. Let's break this down step by step.

### Step 1: Understand the efficiency of a Carnot engine
The efficiency (η) of a Carnot engine is given by the formula:
\[ \eta = 1 - \frac{T_L}{T_H} \]
where \(T_L\) is the absolute temperature of the cold reservoir and \(T_H\) is the absolute temperature of the hot reservoir.

### Step 2: Set up the initial conditions
We are given that the initial efficiency of the Carnot engine is:
\[ \eta = 0.40 \]
Using the efficiency formula:
\[ 0.40 = 1 - \frac{T_L}{T_H} \]
Rearranging this gives:
\[ \frac{T_L}{T_H} = 1 - 0.40 = 0.60 \]
This is our first equation:
\[ T_L = 0.60 T_H \quad \text{(Equation 1)} \]

### Step 3: Apply the changes to the temperatures
According to the problem:
- The temperature of the hot reservoir is quadrupled: \(T_H' = 4T_H\)
- The temperature of the cold reservoir is doubled: \(T_L' = 2T_L\)

### Step 4: Substitute the new temperatures into the efficiency formula
Now we can find the new efficiency using the new temperatures:
\[ \eta' = 1 - \frac{T_L'}{T_H'} \]
Substituting the new temperatures:
\[ \eta' = 1 - \frac{2T_L}{4T_H} \]
This simplifies to:
\[ \eta' = 1 - \frac{1}{2} \cdot \frac{T_L}{T_H} \]

### Step 5: Substitute the value of \(\frac{T_L}{T_H}\)
From Equation 1, we know that \(\frac{T_L}{T_H} = 0.60\). Therefore:
\[ \eta' = 1 - \frac{1}{2} \cdot 0.60 \]
Calculating this gives:
\[ \eta' = 1 - 0.30 = 0.70 \]

### Conclusion
The new efficiency of the Carnot engine after the changes is:
\[ \eta' = 0.70 \]