What is the total amount of heat released when 94.0 g of water at 80.0oC cools to form ice at -30.0oC ?
c(water) = 4.184 J/goC
c(ice) = 2.09 J/goC
To calculate the total amount of heat released, we need to consider two processes: the heating of water from 80.0oC to the boiling point and the cooling of water vapor from the boiling point to -30.0oC.
First, let's calculate the heat required to heat the water from 80.0oC to the boiling point (100.0oC).
To do this, we can use the formula:
Q = m * c * ΔT
where:
Q is the heat absorbed/released (in Joules),
m is the mass of the substance (in grams),
c is the specific heat capacity (in J/goC), and
ΔT is the change in temperature (in oC).
Plugging in the values:
m = 94.0 g
c(water) = 4.184 J/goC
ΔT = 100.0oC - 80.0oC = 20.0oC
Q1 = (94.0 g) * (4.184 J/goC) * (20.0oC)
Next, we need to calculate the heat released during the phase change from water vapor to liquid water.
The heat released during this phase change is given by the formula:
Q2 = m * ΔHf
where:
Q2 is the heat absorbed/released (in Joules),
m is the mass of the substance (in grams), and
ΔHf is the enthalpy of fusion (in J/g).
Since the water is cooling and forming ice, the value of ΔHf should be negative.
Plugging in the values:
m = 94.0 g
ΔHf = -334 J/g (this is the enthalpy of fusion for water, which represents the heat released when 1 gram of water freezes)
Q2 = (94.0 g) * (-334 J/g)
Finally, let's calculate the heat required to cool the ice from 0oC to -30.0oC.
To do this, we can again use the formula:
Q = m * c * ΔT
Plugging in the values:
m = 94.0 g
c(ice) = 2.09 J/goC
ΔT = -30.0oC
Q3 = (94.0 g) * (2.09 J/goC) * (-30.0oC)
To find the total heat released, we need to sum up the three quantities:
Total heat released = Q1 + Q2 + Q3
So, calculate the values for Q1, Q2, and Q3 using the given formulas, and then add them together to find the total heat released.