A Carnot engine whose low-temperature reservoir is at 350 K has efficiency of 50%. It is desired to increase this to 60%. It the temperature of the low-temperature reservoir remains constant, then the temperature reservoir remains constant, then the temperature of the high-temperature reservoir must be increased by how many degrees?

Options

A. 15
B. 175
C. 100
D. 50

Answer

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

\[ \eta = 1 - \frac{T_C}{T_H} \]

where:
- \(\eta\) is the efficiency,
- \(T_C\) is the temperature of the cold reservoir,
- \(T_H\) is the temperature of the hot reservoir.

### Step 1: Calculate the initial temperature of the hot reservoir

Given:
- Efficiency (\(\eta_1\)) = 50% = 0.50
- Temperature of the cold reservoir (\(T_C\)) = 350 K

Using the efficiency formula:

\[ 0.50 = 1 - \frac{350}{T_H} \]

Rearranging the equation:

\[ \frac{350}{T_H} = 1 - 0.50 = 0.50 \]

Now, solving for \(T_H\):

\[ T_H = \frac{350}{0.50} = 700 \text{ K} \]

### Step 2: Calculate the new temperature of the hot reservoir for 60% efficiency

Now, we want to find the new temperature of the hot reservoir (\(T'_H\)) when the efficiency is increased to 60%:

Given:
- New Efficiency (\(\eta_2\)) = 60% = 0.60

Using the efficiency formula again:

\[ 0.60 = 1 - \frac{350}{T'_H} \]

Rearranging the equation:

\[ \frac{350}{T'_H} = 1 - 0.60 = 0.40 \]

Now, solving for \(T'_H\):

\[ T'_H = \frac{350}{0.40} = 875 \text{ K} \]

### Step 3: Calculate the increase in temperature of the hot reservoir

Now, we need to find the increase in temperature:

\[ \Delta T = T'_H - T_H = 875 \text{ K} - 700 \text{ K} = 175 \text{ K} \]

### Final Answer

The temperature of the high-temperature reservoir must be increased by **175 K**.

---