Question

Bananas are to be cooled from 24 to \(13^{\circ} \mathrm{C}\) at a rate of \(215 \mathrm{kg} / \mathrm{h}\) by a refrigeration system. The power input to the refrigerator is \(1.4 \mathrm{kW}\). Determine the rate of cooling, in \(\mathrm{kJ} /\) \(\min ,\) and the COP of the refrigerator. The specific heat of banana above freezing is \(3.35 \mathrm{kJ} / \mathrm{kg} \cdot^{\circ} \mathrm{C}\)

Short answer

Answer: The rate of cooling is approximately 131.55 kJ/min, and the COP of the refrigerator is approximately 1.57.

Step-by-step solution

Calculate the heat removed from the bananas per hour

We are given that the rate of cooling is 215 kg/h. The specific heat of bananas is given as 3.35 kJ/kg·°C, and the change in temperature is (24 - 13) °C. To find the heat removed from the bananas per hour, we can use the formula: Heat removed (Q) = mass × specific heat × change in temperature Q = 215 kg/h × 3.35 kJ/kg·°C × (24°C - 13°C) Q = 215 kg/h × 3.35 kJ/kg·°C × 11°C

Calculate the cooling load in kJ/h

Now, let's find the cooling load by multiplying the mass, specific heat, and the change in temperature. Cooling load = 215 kg/h × 3.35 kJ/kg·°C × 11°C Cooling load = 7893.25 kJ/h

Convert the cooling load to kJ/min

To convert the cooling load from kJ/h to kJ/min, we need to divide it by 60 (as there are 60 minutes in an hour). Cooling load (kJ/min) = 7893.25 kJ/h ÷ 60 min/h Cooling load (kJ/min) = 131.55 kJ/min

Calculate the coefficient of performance (COP)

The COP of a refrigerator is defined as the ratio of the cooling effect (i.e., the cooling load) to the power input. We are given the power input as 1.4 kW. First, we need to convert the power input to kJ/min, knowing that 1 kW = 1000 W and 1 W = 1 J/s. Power input (kJ/min) = 1.4 kW × 1000 W/kW × 60 s/min ÷ 1000 J/kJ Power input (kJ/min) = 84 kJ/min Now, let's find the COP: COP = Cooling load (kJ/min) ÷ Power input (kJ/min) COP = 131.55 kJ/min ÷ 84 kJ/min COP ≈ 1.57 So, the rate of cooling is approximately 131.55 kJ/min, and the COP of the refrigerator is approximately 1.57.