In a Carnot engine, the temperature of the reservoir is `927^(@)C` and that of the skin is `27^(@)C`. The work done by the engine when it transfers heat from the reservoir to the sink is `12.6xx10^(6)J`. What is the quantity of heat absorbed by the engine from the reservoir ?

Options

A. `16.8xx16^(6)J`
B. `4xx16^(6)J`
C. `4.2xx16^(6)J`
D. `7.6xx16^(6)J`

Explanation

To solve the problem, we need to find the quantity of heat absorbed by the Carnot engine from the reservoir, denoted as \( Q_1 \). We are given the following information:

- Temperature of the reservoir (\( T_1 \)): \( 927^\circ C \)
- Temperature of the sink (\( T_2 \)): \( 27^\circ C \)
- Work done by the engine (\( W \)): \( 12.6 \times 10^6 \, J \)

### Step 1: Convert temperatures from Celsius to Kelvin
The absolute temperature in Kelvin is calculated by adding 273 to the Celsius temperature.

\[ T_1 = 927 + 273 = 1200 \, K \]
\[ T_2 = 27 + 273 = 300 \, K \]

### Step 2: Calculate the efficiency of the Carnot engine
The efficiency (\( \eta \)) of a Carnot engine is given by the formula:

\[ \eta = \frac{T_1 - T_2}{T_1} \]

Substituting the values we have:

\[ \eta = \frac{1200 - 300}{1200} = \frac{900}{1200} = \frac{3}{4} \]

### Step 3: Relate efficiency to work done and heat absorbed
The efficiency can also be expressed in terms of work done and heat absorbed:

\[ \eta = \frac{W}{Q_1} \]

From this, we can rearrange to find \( Q_1 \):

\[ Q_1 = \frac{W}{\eta} \]

### Step 4: Substitute the known values into the equation
We know \( W = 12.6 \times 10^6 \, J \) and \( \eta = \frac{3}{4} \):

\[ Q_1 = \frac{12.6 \times 10^6}{\frac{3}{4}} = 12.6 \times 10^6 \times \frac{4}{3} \]

### Step 5: Calculate \( Q_1 \)
Now we calculate \( Q_1 \):

\[ Q_1 = 12.6 \times 10^6 \times \frac{4}{3} = 12.6 \times \frac{4}{3} \times 10^6 \]
\[ Q_1 = 16.8 \times 10^6 \, J \]

### Final Answer
The quantity of heat absorbed by the engine from the reservoir is:

\[ Q_1 = 16.8 \times 10^6 \, J \]

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