How much heat must be removed from 0.20kg of water at 10.0oC to form ice at 0oC?
q1 = heat removed to lower T from 10C to zero C.
q1 = mass H2O x specific heat H2O x (Tfinal-Tinitial)
q1 = 200g x 4.184 x (10) = ?
q2 = heat removed to freeze the H2O (change from liquid to solid phase).
q2 = mass H2O x heat fusion H2O.
q2 = 200 x heat fusion = ?
Look up heat fusion, then total heat removed is q1 + q2
To find out how much heat needs to be removed from water to form ice, we can use the formula:
Q = m * c * ΔT
Where:
Q is the heat energy in Joules
m is the mass of the substance in kilograms
c is the specific heat capacity of the substance in Joules per kilogram per degree Celsius
ΔT is the change in temperature in degrees Celsius.
First, let's calculate the heat energy required to cool the water from 10.0°C to 0.0°C.
The specific heat capacity of water is approximately 4.18 J/g°C or 4180 J/kg°C.
Converting the mass of water from grams to kilograms:
0.20 kg * 1000 g/kg = 200 g
Calculating the heat energy required to cool the water:
Q1 = m * c * ΔT1
= 200 g * 4.18 J/g°C * (0.0°C - 10.0°C)
Note: Since we are cooling the water, the change in temperature (ΔT1) is negative.
Q1 = 200 g * 4.18 J/g°C * (-10.0°C)
= -836 J
Next, let's calculate the heat energy required for the phase change from liquid water to solid ice at 0.0°C.
The specific heat of fusion for water (heat required for phase change) is 334,000 J/kg.
Calculating the heat energy required for the phase change:
Q2 = m * Lf
Where Lf is the specific latent heat of fusion for water.
Q2 = 0.20 kg * 334,000 J/kg
= 66,800 J
To find the total amount of heat energy needed, we add Q1 and Q2:
Q_total = Q1 + Q2
= -836 J + 66,800 J
= 65,964 J
Therefore, approximately 65,964 Joules of heat needs to be removed from 0.20 kg of water at 10.0°C to form ice at 0.0°C.