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 and the given data.

### Step 1: Understand the Efficiency Formula
The efficiency (η) of a Carnot engine is given by the formula:
\[ \eta = 1 - \frac{T_1}{T_2} \]
where:
- \(T_1\) is the temperature of the low-temperature reservoir (in Kelvin),
- \(T_2\) is the temperature of the high-temperature reservoir (in Kelvin).

### Step 2: Identify Given Values
From the problem, we know:
- The low-temperature reservoir \(T_1 = 350 \, K\).
- The initial efficiency \(\eta_i = 50\% = 0.5\).
- The desired efficiency \(\eta_f = 60\% = 0.6\).

### Step 3: Calculate Initial High-Temperature Reservoir \(T_{2i}\)
Using the efficiency formula for the initial condition:
\[ 0.5 = 1 - \frac{350}{T_{2i}} \]
Rearranging gives:
\[ \frac{350}{T_{2i}} = 1 - 0.5 = 0.5 \]
Cross-multiplying:
\[ 350 = 0.5 \cdot T_{2i} \]
Solving for \(T_{2i}\):
\[ T_{2i} = \frac{350}{0.5} = 700 \, K \]

### Step 4: Calculate Final High-Temperature Reservoir \(T_{2f}\)
Now, we apply the efficiency formula for the final condition:
\[ 0.6 = 1 - \frac{350}{T_{2f}} \]
Rearranging gives:
\[ \frac{350}{T_{2f}} = 1 - 0.6 = 0.4 \]
Cross-multiplying:
\[ 350 = 0.4 \cdot T_{2f} \]
Solving for \(T_{2f}\):
\[ T_{2f} = \frac{350}{0.4} = 875 \, K \]

### Step 5: Calculate the Increase in Temperature
To find how much the temperature of the high-temperature reservoir must be increased:
\[ \Delta T = T_{2f} - T_{2i} = 875 \, K - 700 \, K = 175 \, K \]

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

---