Five kg of ice cubes are moved from the freezing compartment of a refrigerator into a home freezer. The refrigerator's compartment is kept at -4.0 celsius. the home freezer is kept at -17 celsius. How much heat does the freezer's cooling systeam remove from the ice cube?
mass of ice x specific heat ice x (Tfinal-Tinitial)
To calculate the heat removed from the ice cube, we can use the formula:
Q = m * c * ΔT
Where:
Q is the heat removed in joules
m is the mass of the ice cube in kilograms
c is the specific heat capacity of ice (approximately 2,090 J/kg·°C)
ΔT is the change in temperature in Celsius
Given:
Mass of the ice cube (m) = 5 kg
Initial temperature (Ti) = -4.0 °C
Final temperature (Tf) = -17 °C
ΔT = Tf - Ti = -17 °C - (-4.0 °C) = -13 °C
Now, let's calculate the heat removed (Q):
Q = 5 kg * 2,090 J/kg·°C * (-13 °C)
Q = -135,850 J
Therefore, the freezer's cooling system removes approximately -135,850 joules of heat from the ice cube. The negative sign indicates that heat is being removed, resulting in a decrease in temperature.
To find the amount of heat removed from the ice cubes, we can use the formula:
Q = m * c * ΔT
Where:
Q is the heat transferred
m is the mass of the object
c is the specific heat capacity of the substance
ΔT is the change in temperature
First, we need to calculate the change in temperature (ΔT). This can be done by subtracting the initial temperature from the final temperature:
ΔT = final temperature - initial temperature
In this case, the initial temperature of the ice cubes is -4.0°C (freezing compartment temperature), and the final temperature is -17°C (home freezer temperature):
ΔT = -17°C - (-4.0°C)
ΔT = -17°C + 4.0°C
ΔT = -13°C
Now we can substitute the values into the formula to calculate the heat transferred (Q). The specific heat capacity of ice is approximately 2.09 J/g°C:
m = 5 kg
c = 2.09 J/g°C
ΔT = -13°C
First, we need to convert the mass from kg to grams:
m = 5 kg * 1000 g/kg
m = 5000 g
Now we can calculate Q:
Q = 5000 g * 2.09 J/g°C * -13°C
Q = -135,850 J
Therefore, the freezer's cooling system removes approximately -135,850 Joules of heat from the ice cubes. The negative sign indicates heat removal.