ice in a refrigerator melts at rate of 3 kg per hour. the temperature outside is 27°c. find the minimum power output of a motor used to drive the refrigerator,which just prevents ice from melting. latent heat of ice = 80 cal/g,J=4.2 joule/cal, 1 h.p = 746 w

To find the minimum power output of the motor used to drive the refrigerator, we need to determine the heat transfer rate from the environment to the refrigerator.

First, let's convert the rate at which ice melts from kg/hr to cal/hr. We know that latent heat of ice is 80 cal/g, so for 3 kg of ice, the heat required to melt it is:

Heat = (80 cal/g) * (3 kg) = 240 cal/hr

Next, let's convert the heat transfer rate from cal/hr to Joules/hr. We know that 1 cal = 4.2 J, so:

Heat = (240 cal/hr) * (4.2 J/cal) = 1008 J/hr

Now, we can calculate the minimum power output of the motor required to transfer this heat away. The power output is the rate at which work is done, which is energy transferred per unit time. Since the time is given in hours, we have:

Power = (1008 J/hr) / (1 hr) = 1008 J/hr

Finally, we can convert the power from Joules/hr to watts (W). We know that 1 W = 1 J/s, and since there are 3600 seconds in an hour:

Power = (1008 J/hr) / (3600 sec/hr) = 0.28 W

Therefore, the minimum power output of the motor used to drive the refrigerator is 0.28 W.

To find the minimum power output of the motor used to drive the refrigerator, we need to consider the heat flow into the refrigerator. The heat flow is given by the formula:

Q = m * L * ΔT

Where:
Q = Heat flow
m = Mass of ice
L = Latent heat of ice
ΔT = Temperature difference

First, we need to convert the given values to the appropriate units:
- Mass of ice: 3 kg (already in SI unit)
- Latent heat of ice: 80 cal/g * 4.2 J/cal = 336 J/g
- Temperature difference: The temperature outside is given as 27°C, and we assume the refrigerator is maintaining the temperature close to freezing point, which is 0°C. So, ΔT = 27 - 0 = 27°C.

Now, let's calculate the heat flow:
Q = 3 kg * 336 J/g * 27°C = 27216 J

To find the power output of the motor, we use the formula:
Power = Energy / Time

Given that the heat flow is occurring over an hour, the time is 1 hour.

Power output = 27216 J / 1 hour = 27216 J / 3600 seconds (since 1 hour = 3600 seconds) = 7.56 W

Finally, to convert the power output to horsepower:
1 horsepower (h.p.) = 746 W
So, the minimum power output of the motor used to drive the refrigerator is:
7.56 W / 746 W/h.p. = 0.01 horsepower

Therefore, the minimum power output of the motor used to drive the refrigerator, which just prevents ice from melting, is 0.01 horsepower.