An electric motor is used to operate a Carnot refrigerator with an interior temperature of 0.00 ◦C. Liquid water at 0.00 ◦C is placed into the refrigerator and transformed to ice at 0.00 ◦C. If the room temperature is 300. K, what mass of ice can be produced in one day by a 0.50 hp motor that is running continuously? Assume that the refrigerator is perfectly insulated and operates at the maximum theoretical efficiency.

732492 g

Explanation:

Given that

Room Temperature = 300 K

Cold T = 0°C = 0 + 273 =273 K

Work done = 0.375 hp

ΔH of fusion of water is 6008 J per mol

The first thing we do is to Convert the power into J/s

Given that 1 hp = 746 W (J/s), then

0.375 hp = 0.375 * 746 W/hp

0.375 = 280 J/s

Then we find the heat per unit time

Q = (cold T / hot T - cold T ) x power

Q = [273 / (300 - 273)] * 280

Q =273/27 * 280

Q = 10.1

Q = 2831 J/s

Moles of ice per second = Q/ ΔH fusion

Moles of ice per second = 2831/6008 = Moles of ice per second = 0.471 mol /s

If we convert 1 day into second, we have

1 day = 1 day * ( 24 hours/day) * ( 60 minutes/hour) * (60 seconds/minutes)

1 * 24 * 3600 = 86400 second

moles of ice /day

0.471 mol / s ) x 86400 s / day =

40694 mols of ice

Molar mass is 18 g/mol

Mass of ice

40694 * 18 = 732492 g

Mass of ice produced per day is 732492 g