A Carnot engine, having an efficiency of `eta= 1/10` as heat engine, is used as a refrigerator. If the work done on the system is 10 J, the amount of energy absorbed from the reservoir at lower temperature is

Options

A. 100J
B. 90J
C. 99J
D. 1J

Answer

To solve the problem, we will follow these steps:

### Step 1: Understand the Carnot Engine and Refrigerator Concept
A Carnot engine operates between two heat reservoirs. It absorbs heat \( Q_1 \) from the lower temperature reservoir and rejects heat \( Q_2 \) to the higher temperature reservoir. The work done by the engine is \( W \).

### Step 2: Use the Efficiency Formula
The efficiency \( \eta \) of a Carnot engine is given by the formula:
\[ \eta = \frac{W}{Q_2} \]
where \( W \) is the work done by the engine and \( Q_2 \) is the heat rejected to the higher temperature reservoir.

### Step 3: Substitute the Given Values
From the problem, we know:
- Efficiency \( \eta = \frac{1}{10} \)
- Work done \( W = 10 \, \text{J} \)

Substituting these values into the efficiency formula:
\[ \frac{1}{10} = \frac{10}{Q_2} \]

### Step 4: Solve for \( Q_2 \)
Rearranging the equation to solve for \( Q_2 \):
\[ Q_2 = 10 \times 10 = 100 \, \text{J} \]

### Step 5: Use the First Law of Thermodynamics
According to the first law of thermodynamics for the Carnot engine:
\[ Q_2 = Q_1 + W \]
We can rearrange this to find \( Q_1 \):
\[ Q_1 = Q_2 - W \]

### Step 6: Substitute Known Values
Now substitute the values of \( Q_2 \) and \( W \):
\[ Q_1 = 100 \, \text{J} - 10 \, \text{J} = 90 \, \text{J} \]

### Final Answer
The amount of energy absorbed from the reservoir at lower temperature is:
\[ \boxed{90 \, \text{J}} \]