A freezer contains 20 kg of food with a specific heat of 2 x 10^3 J/kgC. The freezer motor then operates for 10 minutes reducing the temperature to -8 degrees Celsius.
a) How much heat is removed from the food during this time? The freezer motor operates at 400 watts.
b) How much energy is delivered to the freezer motor during the 10 minute period?
c) During this time, how much total heat is ejected into the room in which the freezer is located?
d) Determine the temperature change in the room if the specific heat of air is 700 J/kgC. Assume there are 80 kg of air in the room, the volume of the air is constant, and there is no heat loss from the room.
I will be happy to critique your thinking.
(a) Q1 =(sp,heat)(grams)(T1-T2)
I think the initial temperature, T1, should be given. As it stands, we have two unknowns, Q1 and T1.
(b) Q2 = (watts)(seconds)
(c) Q3 = Q2-Q1
(d) Q3 = (sp.heat of air)(grams air)(∆T)
Substitute values from previous steps, solve for (∆T)
a) To calculate the heat removed from the food, we can use the formula:
Q = mcΔT
where Q is the heat removed, m is the mass of the food, c is the specific heat, and ΔT is the change in temperature.
Given:
m = 20 kg
c = 2 x 10^3 J/kgC
ΔT = -8°C
Substituting the values into the formula:
Q = (20 kg)(2 x 10^3 J/kgC)(-8°C)
= -320,000 J
Therefore, 320,000 J of heat is removed from the food during this time.
b) The energy delivered to the freezer motor can be calculated using the formula:
Energy = Power × Time
Given:
Power = 400 watts
Time = 10 minutes = 10 × 60 seconds = 600 seconds
Substituting the values into the formula:
Energy = 400 watts × 600 seconds
= 240,000 J
Therefore, 240,000 J of energy is delivered to the freezer motor during the 10-minute period.
c) The total heat ejected into the room can be calculated by subtracting the heat removed from the food (Q) from the energy delivered to the freezer motor.
Total heat ejected = Energy delivered - Heat removed
= 240,000 J - 320,000 J
= -80,000 J
Therefore, 80,000 J of total heat is ejected into the room.
d) To determine the temperature change in the room, we can use the formula:
Q = mcΔT
where Q is the heat transferred, m is the mass of the air, c is the specific heat of air, and ΔT is the change in temperature.
We have already calculated the heat ejected into the room as -80,000 J (negative because it is heat ejected), and we are given the following additional information:
m = 80 kg (mass of air)
c = 700 J/kgC (specific heat of air)
Substituting the values into the formula:
-80,000 J = (80 kg)(700 J/kgC)ΔT
Dividing both sides by (80 kg)(700 J/kgC):
-80,000 J / (80 kg)(700 J/kgC) = ΔT
= -1.43°C
Therefore, the temperature change in the room is -1.43°C.
To solve these problems, we'll use the formula Q = mcΔT, where Q represents the heat energy, m is the mass (in kg), c is the specific heat capacity (in J/kg°C), and ΔT is the change in temperature (in °C).
Let's go through each question step by step:
a) How much heat is removed from the food during this time?
To find the heat removed from the food, we need to calculate the change in temperature first. The initial temperature of the food is not mentioned, but assuming it's at 0°C (since it's in a freezer), the change in temperature is (0°C - (-8°C)) = 8°C.
Now, we can calculate the heat removed using the formula mentioned above:
Q = mcΔT
Q = (20 kg) * (2 x 10^3 J/kg°C) * (8°C)
Q = 320,000 J
Therefore, 320,000 J of heat is removed from the food during this time.
b) How much energy is delivered to the freezer motor during the 10-minute period?
To find the energy delivered to the freezer motor, we can use the formula Power (P) = Energy (E) / Time (t), where the power is given as 400 watts, and the time is 10 minutes. However, we need to convert the time into seconds since power is given in watts (J/s).
t = 10 minutes * (60 seconds/1 minute)
t = 600 seconds
Now we can calculate the energy delivered:
P = E / t
E = P * t
E = 400 watts * 600 seconds
E = 240,000 J
Therefore, 240,000 J of energy is delivered to the freezer motor during the 10-minute period.
c) During this time, how much total heat is ejected into the room in which the freezer is located?
Since the freezer removes heat from the food, that heat is transferred into the room. Therefore, the total heat ejected into the room is equal to the heat removed from the food.
From part a, we found that the heat removed from the food is 320,000 J. So, during this time, 320,000 J of heat is ejected into the room.
d) Determine the temperature change in the room.
To find the temperature change in the room, we'll use the same formula:
Q = mcΔT
Here, we need to calculate for ΔT. The mass and specific heat capacity of the air are given, and the heat ejected into the room is also calculated in part c.
Q = mcΔT
320,000 J = (80 kg) * (700 J/kg°C) * ΔT
Now, we can solve for ΔT:
ΔT = 320,000 J / (80 kg * 700 J/kg°C)
ΔT ≈ 5 °C
Therefore, the temperature in the room increases by approximately 5 °C.
Note: Since ΔT is positive, it means the temperature of the room increases.