Farmers often spray water on plants when there is a chance of frost to protect them from freezing. Calculate the amount of heat released when 100 mL of water at 25oC cools to -2oC. (ΔHfusion water is 6.03 kJ/mol, cwater = 4.18 J/(goC), cice = 2.01 J/(goC))
I worked this for you at your earlier post.
To calculate the amount of heat released, we need to consider the heat gained by the water as it cools down and then the heat lost during the freezing of water into ice.
First, let's calculate the heat gained by water as it cools down from 25°C to 0°C using the specific heat capacity equation:
q1 = m * c * ΔT
Where:
q1 = heat gained or lost
m = mass of the substance (water)
c = specific heat capacity of the substance (water)
ΔT = change in temperature
Given:
m = 100 mL (since the density of water is 1 g/mL, the volume is equal to mass)
c = 4.18 J/(g·°C)
ΔT = 0°C - 25°C = -25°C
Let's substitute the values into the equation:
q1 = (100 g) * (4.18 J/(g·°C)) * (-25°C)
q1 = -10450 J
Next, we need to calculate the heat released during the freezing of water into ice. We can use the enthalpy of fusion of water:
ΔHfusion = q2
Where:
ΔHfusion = enthalpy of fusion of water
q2 = heat gained or lost
Given:
ΔHfusion = 6.03 kJ/mol (kilojoules per mole)
Since we have 100 mL of water, we can convert it to moles using the molar mass of water (18.015 g/mol):
mol = (100 g) / (18.015 g/mol)
mol = 5.55 mol
Now we can calculate q2:
q2 = (5.55 mol) * (6.03 kJ/mol) * (1000 J/kJ)
q2 = 33353.5 J
The total heat released is the sum of q1 and q2:
total heat released = q1 + q2
total heat released = -10450 J + 33353.5 J
total heat released = 22903.5 J
Therefore, the amount of heat released when 100 mL of water at 25°C cools to -2°C is 22903.5 J.