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))
q1 = heat released on cooling liquid water from 25C to zero C.
q1 = mass H2O x specfic heat water x (Tfinal-Tinitial)---for example, Tf is zero and Ti is 25)
q2 = heat released on freezing liquid water at zero C to ice at zero C.
q2 = mass water x heat fusion.
q3 = heat released on cooling ice at zero C to -2 C.
q3 = mass ice x specific heat ice x (Tfinal-Tinitial)
Total heat released is q1 + q2 + q3.
In q3 (Tfinal-Tinitial) is Tf = -2 and Ti = zero ?
Yes, Tf is -2 and Ti is zero.
To calculate the amount of heat released when 100 mL of water cools from 25oC to -2oC, we need to determine the heat lost during the cooling process.
The first step is to calculate the mass of water in grams using the density of water, which is approximately 1 g/mL:
Mass of water = volume × density = 100 mL × 1 g/mL = 100 g
The next step is to calculate the heat lost during the cooling process. We can break the process into two parts: the heat lost when water cools from 25oC to 0oC and the heat lost when water freezes from 0oC to -2oC.
1. Heat lost from cooling water from 25oC to 0oC:
ΔQ1 = mass × specific heat capacity × temperature change
ΔQ1 = 100 g × 4.18 J/(g⋅°C) × (0°C – 25°C)
ΔQ1 = -10,450 J
2. Heat lost from water freezing at 0oC to -2oC:
Since water freezes at 0oC, heat is released during the phase change from liquid to solid. The heat released during this phase change is known as the heat of fusion.
ΔHfusion = 6.03 kJ/mol = 6.03 × 10^3 J/mol (1 kJ = 10^3 J)
First, we need to convert the mass of water in grams to moles:
Number of moles = mass / molar mass
Molar mass of water = 2(1.01 g/mol of hydrogen) + 16.00 g/mol of oxygen = 18.02 g/mol
Number of moles = 100 g / 18.02 g/mol ≈ 5.55 mol
Now we can calculate the heat lost during freezing:
ΔQ2 = ΔHfusion × moles of water
ΔQ2 = 6.03 × 10^3 J/mol × 5.55 mol
ΔQ2 = 33,416.5 J
Now, we can calculate the total heat lost during the entire cooling process:
Total heat lost = ΔQ1 + ΔQ2
Total heat lost = -10,450 J + 33,416.5 J
Total heat lost = 22,966.5 J
Therefore, the amount of heat released when 100 mL of water at 25oC cools to -2oC is approximately 22,966.5 Joules.