How much energy (in kilojoules) is needed to heat 4.00g of ice from -11.0∘C to 21.5∘C? The heat of fusion of water is 6.01kJ/mol, and the molar heat capacity is 36.6 J/(K⋅mol) for ice and 75.3 J/(K⋅mol) for liquid water.
Three equations will work most of the heat problems.
When you are heating WITHIN a phase (raising a liquid @ one T to liquid at another T; solid at one T to solid at another T; gas at one T to gas at another T) use
q = mass x specific heat in that phase x (Tfinal-Tintitial)
When you are at a phase change use one of these two.
For phase change from solid to liquid (the melting point is where you change from solid phase to liquid phase) use
q = mass x heat fusion.
For phase change from liquid to gas (the boiling point is where you change from the liquid phase to the gas phase) use
q = mass x heat vaporization.
For the above problem you have
solid ice moving from -11 C to zero C (that's within a phase).
Then you have ice melting (phase change)
Then you have liquid water going from zero C to 21.5 C.(another within a phase)
Total q is the sum of q1 + q2 + q3
To calculate the energy required to heat the ice, we need to consider two steps:
1. The energy required to raise the temperature of the ice from -11.0∘C to 0∘C.
2. The energy required to melt the ice at 0∘C to liquid water at 0∘C.
3. The energy required to raise the temperature of the water from 0∘C to 21.5∘C.
Let's break down each step:
1. Energy required to raise the temperature of the ice from -11.0∘C to 0∘C:
We can use the formula Q = m * c * ΔT, where Q is the energy, m is the mass, c is the specific heat capacity, and ΔT is the change in temperature. Since the specific heat capacity of ice is given in J/(K⋅mol), we need to convert it to J/(g⋅K).
First, convert the mass of ice from grams to moles by dividing by the molar mass of water (18.015 grams/mole).
moles = mass / molar mass
= 4.00g / 18.015 g/mol
= 0.2221 mol
Next, convert the specific heat of ice from J/(K⋅mol) to J/(g⋅K):
specific heat (J/(g⋅K)) = specific heat (J/(K⋅mol)) / molar mass (g/mol)
= 36.6 J/(K⋅mol) / 18.015 g/mol
= 2.031 J/(g⋅K)
Now we can calculate the energy using ΔT = 0∘C - (-11.0∘C) = 11.0∘C:
Q1 = m * c * ΔT
= 0.2221 mol * 2.031 J/(g⋅K) * 11.0∘C
= 5.058 J
Since we have values in kilojoules, we can convert J to kJ:
Q1 = 5.058 J / 1000
≈ 0.0051 kJ
2. Energy required to melt the ice at 0∘C to liquid water at 0∘C:
The heat of fusion of water, which is the energy required to change a substance from a solid to a liquid phase at its melting point, is given as 6.01 kJ/mol.
We already calculated the number of moles of water from the mass of ice:
moles = 0.2221 mol
Therefore, the energy required to melt the ice is:
Q2 = moles * heat of fusion
= 0.2221 mol * 6.01 kJ/mol
= 1.334 kJ
3. Energy required to raise the temperature of the water from 0∘C to 21.5∘C:
Now we need to consider liquid water. The specific heat capacity for liquid water is 75.3 J/(K⋅mol). We already calculated the moles of water, which is still 0.2221 mol. The change in temperature is ΔT = 21.5∘C - 0∘C = 21.5∘C.
Q3 = moles * c * ΔT
= 0.2221 mol * 75.3 J/(K⋅mol) * 21.5∘C
= 329.997 J
Again, we convert J to kJ:
Q3 = 329.997 J / 1000
≈ 0.330 kJ
To find the total energy required, we sum up the three parts:
Total energy = Q1 + Q2 + Q3
≈ 0.0051 kJ + 1.334 kJ + 0.330 kJ
≈ 1.669 kJ
Therefore, approximately 1.669 kilojoules of energy are needed to heat 4.00 grams of ice from -11.0∘C to 21.5∘C.