How much energy (in kilojoules) is needed to heat 5.55g of ice from -13.5 ∘ C to 26.5 ∘ 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.
So this in steps.
Within a phase the formula is
q = mass x specific heat x (Tfinal-Tinitial).
For example, to heat liquid H2O from zero C to 100 C use
q = mass x specific heat x (100-0) = ?
At a phase change (from ice to liquid at the melting point or from liquid to steam at 100 c), use
q = mass x heat fusion (at melting point) or
q = mass x heat vaporization (at boiling point)
Then add each q to find the sum.
To find the amount of energy needed to heat the ice, we need to consider two stages: first, we need to heat the ice from -13.5 °C to 0 °C, and then we need to melt the ice into liquid water at 0 °C, and finally, we need to heat the water from 0 °C to 26.5 °C.
Let's break down the calculation step by step:
1. Calculate the energy required to heat the ice from -13.5 °C to 0 °C:
The molar heat capacity for ice is given as 36.6 J/(K⋅mol).
The molar mass of water is approximately 18 g/mol.
We need to convert grams of ice to moles:
Moles of ice = (mass of ice)/(molar mass of water)
= 5.55 g / 18 g/mol
≈ 0.30833 mol
Energy required = moles of ice * heat capacity * temperature change
= 0.30833 mol * 36.6 J/(K⋅mol) * (0 - (-13.5))
= 0.30833 mol * 36.6 J/(K⋅mol) * 13.5 K
≈ 169.13 J
Since 1 kJ = 1000 J, the energy required to heat the ice from -13.5 °C to 0 °C is approximately:
169.13 J / 1000
≈ 0.16913 kJ
2. Calculate the energy required to melt the ice at 0 °C:
The heat of fusion (enthalpy of fusion) for water is given as 6.01 kJ/mol.
Energy required = moles of ice * heat of fusion
= 0.30833 mol * 6.01 kJ/mol
≈ 1.8493 kJ
3. Calculate the energy required to heat the liquid water from 0 °C to 26.5 °C:
The molar heat capacity for liquid water is given as 75.3 J/(K⋅mol).
Energy required = moles of water * heat capacity * temperature change
= 0.30833 mol * 75.3 J/(K⋅mol) * (26.5 - 0)
= 0.30833 mol * 75.3 J/(K⋅mol) * 26.5 K
≈ 613.49 J
Since 1 kJ = 1000 J, the energy required to heat the water from 0 °C to 26.5 °C is approximately:
613.49 J / 1000
≈ 0.61349 kJ
Now, to find the total energy required, we add up the energy calculated in each step:
Total energy = energy to heat from -13.5 °C to 0 °C + energy to melt the ice + energy to heat from 0 °C to 26.5 °C
≈ 0.16913 kJ + 1.8493 kJ + 0.61349 kJ
≈ 2.63192 kJ
Therefore, approximately 2.63192 kilojoules of energy is needed to heat 5.55 grams of ice from -13.5 °C to 26.5 °C.