0.50 L of water at 16°C is poured into an aluminum ice cube tray of mass 0.250 kg at the same temperature. How much energy must be removed from this system by the refrigerator to turn the water into ice at -8°C?
Specific heat capacity for aluminum = 920 Jkg−1K−1 Specificheatcapacityforice=2100 Jkg−1K−1
Latent heat for water = 3.33 × 105 Jkg−1
To determine the amount of energy that needs to be removed from the system to turn the water into ice, we can use the equation:
Q = mcΔT
Where:
Q is the amount of energy transferred
m is the mass of the substance
c is the specific heat capacity of the substance
ΔT is the change in temperature
Step 1: Calculate the energy required to lower the temperature of the water from 16°C to 0°C
First, we need to calculate the energy required to lower the temperature of the water from 16°C to 0°C.
Q1 = mcΔT
= (0.50 kg) * (4.18 kJ/kg°C) * (0°C - 16°C)
= -33.44 kJ
Step 2: Calculate the energy required to freeze the water at 0°C
Next, we need to calculate the energy required to freeze the water at 0°C.
Q2 = mL
= (0.50 kg) * (334 kJ/kg)
= 167 kJ
Step 3: Calculate the energy required to lower the temperature of the ice from 0°C to -8°C
Finally, we need to calculate the energy required to lower the temperature of the ice from 0°C to -8°C.
Q3 = mcΔT
= (0.25 kg) * (0.897 kJ/kg°C) * (-8°C - 0°C)
= -1.79 kJ
Step 4: Calculate the total energy required
To determine the total energy required, we sum up Q1, Q2, and Q3.
Q_total = Q1 + Q2 + Q3
= -33.44 kJ + 167 kJ - 1.79 kJ
= 131.77 kJ
Therefore, the refrigerator needs to remove 131.77 kJ of energy from the system to turn the water into ice at -8°C.
Three steps
Q = mc(Tf-Ti) to get 16 to zero for both water and aluminum
Q = mL to turn water to ice
Q = mc(Tf-Ti) to get zero to -8 for both water and aluminum.
Don't know specific heats of water, ice or aluminum offhand and I don't know latent heat of fusion for water)