Question

A container holding \(1.20 \mathrm{kg}\) of water at \(20.0^{\circ} \mathrm{C}\) is placed in a freezer that is kept at \(-20.0^{\circ} \mathrm{C} .\) The water freezes and comes to thermal equilibrium with the interior of the freezer. What is the minimum amount of electrical energy required by the freezer to do this if it operates between reservoirs at temperatures of $20.0^{\circ} \mathrm{C}\( and \)-20.0^{\circ} \mathrm{C} ?$

Short answer

Answer: The minimum amount of electrical energy required is approximately 3,317,647 J.

Step-by-step solution

Calculate the Carnot efficiency

The Carnot efficiency is given by the formula: $$\eta_{Carnot} = 1 - \frac{T_{cold}}{T_{hot}}$$ where \(T_{cold}\) and \(T_{hot}\) are absolute temperatures, in Kelvin, of the cold and hot reservoirs, respectively. First, we need to convert Celsius temperatures to Kelvin, then we will plug the values into the formula.

Convert temperatures to Kelvin

To convert Celsius temperatures to Kelvin, we can use the following relation: $$T_{K} = T_{C} + 273.15$$ So, the temperatures in Kelvin are $$T_{cold} = -20.0^{\circ}\mathrm{C} + 273.15 = 253.15 \mathrm{K}$$ $$T_{hot} = 20.0^{\circ}\mathrm{C} + 273.15 = 293.15 \mathrm{K}$$

Compute the Carnot efficiency

Now, we can plug the temperatures into the Carnot efficiency formula: $$\eta_{Carnot} = 1 - \frac{T_{cold}}{T_{hot}} = 1 - \frac{253.15 \mathrm{K}}{293.15 \mathrm{K}} \approx 0.136$$ So, the Carnot efficiency is approximately \(13.6 \%\).

Calculate the heat removed from the water

We need to calculate the amount of heat removed from the water, which consists of two parts: the heat removed to freeze the water, and the heat removed to further cool the ice down to \(-20.0^{\circ}\mathrm{C}\). First, we need to calculate the heat removed to freeze the water using the specific latent heat of fusion \(L_f\): $$Q_{fusion} = m \cdot L_f$$ For water, \(L_f = 334\,000 \, \mathrm{J/kg}\). With \(m = 1.20\, \mathrm{kg}\), we have: $$Q_{fusion} = 1.20\, \mathrm{kg} \cdot 334\,000\, \mathrm{J/kg} = 400\,800\, \mathrm{J}$$ Next, we will calculate the heat removed to cool the ice down to \(-20.0^{\circ}\mathrm{C}\) using the specific heat capacity \(c_{ice}\): $$Q_{cooling} = m \cdot c_{ice} \cdot \Delta T$$ For ice, \(c_{ice} = 2\,100\, \mathrm{J/(kg\cdot K)}\). With \(\Delta T = 20.0\, \mathrm{K}\), we have: $$Q_{cooling} = 1.20\, \mathrm{kg} \cdot 2\,100\, \mathrm{J/(kg\cdot K)} \cdot 20.0\, \mathrm{K} = 50\,400\, \mathrm{J}$$ Now we can find the total heat removed: $$Q_{total} = Q_{fusion} + Q_{cooling} = 400\,800\, \mathrm{J} + 50\,400\, \mathrm{J} = 451\,200\, \mathrm{J}$$

Calculate the minimum electrical energy required

Finally, we can calculate the minimum electrical energy needed, using the Carnot efficiency: $$W_{min} = \frac{Q_{total}}{\eta_{Carnot}} = \frac{451\,200\, \mathrm{J}}{0.136} \approx 3\,317\,647\, \mathrm{J}$$ Therefore, the minimum amount of electrical energy required by the freezer to freeze \(1.20\, \mathrm{kg}\) of water at \(20.0^{\circ}\mathrm{C}\) and bring it to thermal equilibrium with the interior at \(-20.0^{\circ}\mathrm{C}\) is approximately \(3\,317\,647\, \mathrm{J}\).