A Carnot engine whose low temperature reservoir is at `7^(@)C` has an efficiency of `50%`. It is desired to increase the efficiency to `70%` By low many degrees should the temperature of the high temperature reservoir be increased

Options

A. `840 K`
B. `280 K`
C. `560 K`
D. `380 K`

Explanation

To solve the problem, we will follow these steps:

### Step 1: Convert the low temperature reservoir from Celsius to Kelvin
The low temperature reservoir (T2) is given as \(7^\circ C\). To convert this to Kelvin, we use the formula:
\[ T(K) = T(°C) + 273 \]
So,
\[ T2 = 7 + 273 = 280 \, K \]

### Step 2: Use the efficiency formula to find the initial high temperature reservoir (T1)
The efficiency (\(η\)) of a Carnot engine is given by:
\[ η = 1 - \frac{T2}{T1} \]
Given that the initial efficiency is \(50\%\) or \(0.5\):
\[ 0.5 = 1 - \frac{280}{T1} \]
Rearranging gives:
\[ \frac{280}{T1} = 0.5 \implies T1 = \frac{280}{0.5} = 560 \, K \]

### Step 3: Calculate the new high temperature reservoir for the desired efficiency
Now we want to increase the efficiency to \(70\%\) or \(0.7\):
\[ 0.7 = 1 - \frac{T2}{T1'} \]
Where \(T1'\) is the new high temperature. Rearranging gives:
\[ \frac{T2}{T1'} = 0.3 \implies T1' = \frac{T2}{0.3} = \frac{280}{0.3} \approx 933.33 \, K \]

### Step 4: Calculate the increase in temperature of the high temperature reservoir
Now we find the increase in temperature:
\[ \Delta T = T1' - T1 = 933.33 - 560 = 373.33 \, K \]

### Final Answer
The high temperature reservoir should be increased by approximately \(373.33 \, K\).

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