Answer:
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