A Carnot engine having an efficiency of $$\frac{1}{10}$$ is being used as a refrigerator. If the work done on the refrigerator is 10 J, the amount of heat absorbed from the reservoir at a lower temperature is

We have a Carnot engine whose efficiency when it works as a heat engine is given as $$\dfrac{1}{10}$$. First, recall the definition of efficiency for a Carnot heat engine:

$$\eta \;=\;\dfrac{W}{Q_h}\;=\;1-\dfrac{T_c}{T_h}$$

Here $$\eta$$ is the efficiency, $$W$$ is the work obtained from the engine, $$Q_h$$ is the heat absorbed from the high-temperature reservoir, $$T_h$$ is the absolute temperature of the hot reservoir, and $$T_c$$ is the absolute temperature of the cold reservoir.

Substituting the given efficiency, we write

$$\eta \;=\;\dfrac{1}{10}\;=\;1-\dfrac{T_c}{T_h}.$$

Now we isolate the temperature ratio. Moving the fraction to the right-hand side and the constant term to the left, we obtain

$$\dfrac{T_c}{T_h}\;=\;1-\dfrac{1}{10}.$$

The right side simplifies as

$$1-\dfrac{1}{10}\;=\;\dfrac{10}{10}-\dfrac{1}{10}\;=\;\dfrac{9}{10}.$$

So,

$$\dfrac{T_c}{T_h}\;=\;\dfrac{9}{10}=0.9.$$

Next, we change the mode of operation: the same Carnot device is now used as a refrigerator. For a Carnot refrigerator, the coefficient of performance (COP) is defined by the formula

$$\text{COP}_{\text{ref}}\;=\;\dfrac{Q_c}{W}\;=\;\dfrac{T_c}{T_h-T_c},$$

where $$Q_c$$ is the heat drawn from the cold (low-temperature) reservoir and $$W$$ is the work input to run the refrigerator.

We already know $$\dfrac{T_c}{T_h}=0.9$$, so let us express $$T_c$$ and $$T_h-T_c$$ in terms of $$T_h$$:

$$T_c = 0.9\,T_h,$$

$$T_h - T_c = T_h - 0.9\,T_h = 0.1\,T_h.$$

Substituting these values into the COP expression, we get

$$\text{COP}_{\text{ref}} = \dfrac{0.9\,T_h}{0.1\,T_h}.$$

The common factor $$T_h$$ cancels out, leaving

$$\text{COP}_{\text{ref}} = \dfrac{0.9}{0.1}=9.$$

This means that for every 1 joule of work supplied to the refrigerator, it absorbs 9 joules of heat from the low-temperature reservoir.

Now, the problem states that the work done on the refrigerator is 10 J. Using the definition of COP once again,

$$\text{COP}_{\text{ref}} = \dfrac{Q_c}{W}.$$

Substituting $$\text{COP}_{\text{ref}} = 9$$ and $$W = 10\;\text{J}$$, we find

$$9 = \dfrac{Q_c}{10\;\text{J}}.$$

Multiplying both sides by 10 J, we get

$$Q_c = 9 \times 10\;\text{J} = 90\;\text{J}.$$

So, the refrigerator absorbs 90 J of heat from the low-temperature reservoir.

Hence, the correct answer is Option D.