In an electric freezer, 400.0 g of water at 18.0 degree C is cooled, frozen, and the ice is chilled to -5.00 degree C. How much total heat was removed from the water
To calculate the total heat removed from the water, we need to consider the heating/cooling processes separately.
1. Heating water from 18.0°C to 0°C:
The specific heat capacity of water is 4.18 J/g°C.
The formula to calculate the heat (Q) transferred is Q = m * c * ΔT, where m is the mass, c is the specific heat capacity, and ΔT is the temperature change.
Q1 = (400.0 g) * (4.18 J/g°C) * (0°C - 18.0°C)
Q1 = -3012.0 J
2. Melting the ice at 0°C:
The heat of fusion for water is 334 J/g.
The formula to calculate the heat (Q) transferred is Q = m * ΔHf, where m is the mass of ice and ΔHf is the heat of fusion.
Q2 = (400.0 g) * (334 J/g)
Q2 = 133600.0 J
3. Cooling the ice from 0°C to -5.00°C:
The specific heat capacity of ice is 2.09 J/g°C.
The formula to calculate the heat (Q) transferred is Q = m * c * ΔT, where m is the mass, c is the specific heat capacity, and ΔT is the temperature change.
Q3 = (400.0 g) * (2.09 J/g°C) * (-5.00°C - 0°C)
Q3 = -4180.0 J
Therefore, the total heat removed from the water is the sum of Q1, Q2, and Q3:
Total heat removed = Q1 + Q2 + Q3
Total heat removed = -3012.0 J + 133600.0 J - 4180.0 J
Total heat removed = 126408.0 J
So, the total heat removed from the water is 126,408.0 J.
To determine the total heat removed from the water, we need to take into account the change in temperature during the cooling and freezing process.
First, let's calculate the heat required to cool the water from 18.0°C to 0°C. We can use the formula:
Q1 = m * C * ΔT
Where:
Q1 is the heat required to raise the temperature
m is the mass of the water (400.0 g)
C is the specific heat capacity of water (4.18 J/g°C)
ΔT is the change in temperature (from 18.0°C to 0°C)
Substituting the values into the formula, we have:
Q1 = 400.0 g * 4.18 J/g°C * (0°C - 18.0°C)
Simplifying the calculation:
Q1 = 400.0 g * 4.18 J/g°C * (-18.0°C)
Q1 = -30192 J
Since we are removing heat from the system, the value is negative, indicating that heat is being taken out.
Next, let's calculate the heat required to freeze the water at 0°C. We can use the formula:
Q2 = m * ΔHf
Where:
Q2 is the heat required for the phase change (freezing)
m is the mass of the water (400.0 g)
ΔHf is the enthalpy of fusion for water (334 J/g)
Substituting the values into the formula, we have:
Q2 = 400.0 g * 334 J/g
Q2 = 133600 J
Again, since we are removing heat from the system, the value is positive.
Finally, let's calculate the heat required to chill the ice from 0°C to -5.00°C. Since ice's specific heat is different from water's, we'll use a different specific heat capacity value, specifically 2.09 J/g°C. Using the same formula:
Q3 = m * C * ΔT
Q3 = 400.0 g * 2.09 J/g°C * (0°C - (-5.00°C))
Simplifying the calculation:
Q3 = 400.0 g * 2.09 J/g°C * (5.00°C)
Q3 = 4180 J
Again, since we are removing heat from the system, the value is negative.
Finally, to calculate the total heat removed, we add the three values:
Total Heat Removed = Q1 + Q2 + Q3
= -30192 J + 133600 J - 4180 J
= 98928 J
Therefore, the total heat removed from the water is 98,928 J.
heat to cool water to 0C: 400g*cwater*18
heat to make ice at OC: 400*Hf
heat to cool ice down: 400(cice)(5)
look up the specific heats of ice, and water, memorize them. all the heats removed