How much time does it take a freezer to make a tray of ice cubes out of a tray of 0.25kg liquid water at 10 celsius). assume the freezer is a constant cycle operating between-8(celsius) and 35(celsius) with a power requirement of 750 n. assume the liquid water is the only contents in the freezer

time = 20.27 second

Explanation:

Given data

power = 750 N

ice cube = 0.25 kg

liquid water temp = 10°C

cycle operating = -8°C  to 35°C

to find out

required time

solution

we know that latent heat of water i.e. h(w) = 333 kJ/kg

and specific heat of water i.e S(p) = 4.187 kJ/kg

so we find first heat remove from water (Q)  i.e = mass (specific heat to water temp (10°C to  0°C) + latent heat of water)

so Q =  0.25 ( 4.187 (10°C -0°C ) + 333)

Q =  93.7175 kJ

in question we have given

Assume the freezer is a constant cycle operating so we consider here reversible carnot cycle

so Coefficient of Performance = desired effect / work input = heat absorbed  from evaporation / power

so here desire effect is 273 -8 = 265 and work input  is 35 - (-8) = 43

power is 0.75

so heat absorbed from evaporation = 265 / 43 × 0.75

heat absorbed from evaporation = 4.622 kJ/s

and we know heat absorbed from evaporation =  heat remove from water / time

so time will be 93.7175 / 4.622

time = 20.27 second

answer:

try looking it up that is always

explanation: