An ideally efficient freezer cools food to 260 K. If the room temperature is 300K, how much work does the freezer consume when removing 100J heat from the food?
To calculate the work done by the freezer, we will use the first law of thermodynamics, which states that the work done is equal to the difference in heat energy.
The formula is:
Work = Heat In - Heat Out
In this case, since the freezer is removing heat from the food, the heat flow is negative.
Given:
Heat Out = -100 J (heat removed from the food)
Room Temperature = 300 K
Food Temperature = 260 K
To find the heat in, we can use the equation:
Heat In = Heat Out - Cooling
The cooling of the heat refers to the energy required to lower the food temperature from the room temperature to the desired temperature.
The formula for cooling is given by:
Cooling = m * C * ΔT
Where:
m = mass of the food
C = specific heat capacity of the food
ΔT = change in temperature (Room Temperature - Food Temperature)
Since the mass and specific heat capacity of the food are not provided, we won't be able to calculate the exact value of the cooling in this case. However, we can still calculate the work done by the freezer by using the given information.
Given:
Heat Out = -100 J
Room Temperature = 300 K
Food Temperature = 260 K
Using the equation:
Work = Heat In - Heat Out
Work = (-Heat Out) - Cooling
Substituting the given values:
Work = (-(-100 J)) - Cooling
Work = 100 J - Cooling
Hence, the work consumed by the freezer when removing 100 J heat from the food is 100 J minus the cooling required to lower the food temperature. Note that without specific values for mass and specific heat capacity of the food, we cannot determine the exact work done.