A Carnot engine having an efficiency of \(\frac {1}{10}\) 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

Question

Which is an adiabatic process?

Answer 1

To solve this problem, we need to understand the working principles of a Carnot engine and its dual function as a refrigerator. A Carnot engine, when used as a refrigerator, transfers heat from a low-temperature reservoir to a high-temperature reservoir using work supplied to it.

First, given that the efficiency \(\eta\) of the Carnot engine as a heat engine is \(\frac{1}{10}\), we can use the efficiency formula for heat engines:

\(\eta = 1 - \frac{T_C}{T_H}\)

where \(T_C\) is the temperature of the cold reservoir and \(T_H\) is the temperature of the hot reservoir. As a refrigerator, the role changes to determining the Coefficient of Performance (COP), which is defined as:

\(\text{COP} = \frac{Q_C}{W}\)

where \(Q_C\) is the heat absorbed from the cold reservoir and \(W\) is the work done on the system.

The relation between efficiency \(\eta\) and COP for a Carnot refrigerator is:

\(\text{COP} = \frac{1}{\eta} - 1\)

Substituting \(\eta = \frac{1}{10}\), the COP becomes:

\(\text{COP} = \frac{1}{\frac{1}{10}} - 1 = 10 - 1 = 9\)

Now, using the equation for COP:

\(9 = \frac{Q_C}{10}\)

Solving for \(Q_C\), the energy absorbed from the reservoir at lower temperature:

\(Q_C = 9 \times 10 = 90 \, \text{J}\)

Therefore, the amount of energy absorbed from the lower temperature reservoir is 90 J. Thus, the correct answer is Option 2: 90 J.

Answer 2