Answer:
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