Final answer:
To cool 5 watermelons from 28 °C to 8 °C in a refrigerator with a COP of 1.5 and power input of 450 W, it would take approximately 6222.22 seconds, or about 1 hour and 43 minutes.
Step-by-step explanation:
The question asks to determine the time it would take to cool watermelons in a household refrigerator with specific energy input and coefficient of performance (COP). This type of problem involves concepts of thermodynamics and heat transfer, which are areas within the field of physics.
To solve this problem, we need to calculate the total heat that must be removed from the watermelons to lower their temperature from 28 °C to 8 °C. The heat removed (Q) is given by the equation Q = mcΔT, where m is the mass of the watermelons, c is the specific heat of water, and ΔT is the change in temperature.
For 5 watermelons each of 10 kg, the total mass (m) is 50 kg. The specific heat (c) of watermelon, treated as water, is 4.2 kJ/kg.°C. The change in temperature (ΔT) is 28 °C - 8 °C = 20°C. Thus, Q = (50 kg)(4.2 kJ/kg.°C)(20°C) = 4200 kJ.
The refrigerator has a power input of 450 W and a COP of 1.5. The refrigeration effect (Qe) can be calculated as Qe = COP × Power input. Thus, Qe = 1.5 × 450 W = 675 W or 0.675 kJ/s.
To find the time (t) it takes to remove 4200 kJ, we can use the formula t = Q / Qe. Therefore, t = 4200 kJ / 0.675 kJ/s = 6222.22 s.
So, it would take approximately 6222.22 seconds, or roughly 1 hour and 43 minutes, to cool the watermelons to 8 °C.