Question

The \(\mathrm{COP}\) of a refrigerator decreases as the temperature of the refrigerated space is decreased. That is, removing heat from a medium at a very low temperature will require a large work input. Determine the minimum work input required to remove \(1 \mathrm{kJ}\) of heat from liquid helium at \(3 \mathrm{K}\) when the outside temperature is 300 K.

Short answer

Answer: The minimum work input required is approximately 99.01 kJ.

Step-by-step solution

Determine the Minimum Coefficient of Performance for Refrigerator

Using the Carnot efficiency for refrigerators, we can determine the minimum COP as: $$COP_{min} = \frac{T_L}{T_H - T_L}$$ Plugging in the given values \(T_L = 3 K\) and \(T_H = 300 K\), we obtain: $$COP_{min} = \frac{3}{300 - 3}$$

Compute the Minimum Coefficient of Performance

Carrying out the division, we get: $$COP_{min} = \frac{3}{297}$$ $$COP_{min} = 0.0101$$

Use the COP to Determine the Minimum Work Input Required

Now that we have the minimum Coefficient of Performance, we can find the minimum work input required using the formula for COP: $$COP = \frac{Q_L}{W_{in}}$$ Solving for \(W_{in}\): $$W_{in} = \frac{Q_L}{COP}$$ We know that \(Q_L = 1 kJ\) and \(COP = 0.0101\), so we can plug in these values: $$W_{in} = \frac{1}{0.0101}$$

Compute the Minimum Work Input

Now, dividing the heat removed by the minimum COP, we find the minimum work input required: $$W_{in} = 99.01 kJ$$ So, the minimum work input required to remove 1 kJ of heat from liquid helium at 3 K when the outside temperature is 300 K is approximately 99.01 kJ.