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.
Didn't I do this earlier? (or at least one just like it)
To calculate the energy needed to heat the ice, we need to consider two steps: (1) the energy required to raise the temperature of the ice to its melting point, and (2) the energy needed to melt the ice and raise the temperature of the resulting water to the final temperature.
Step 1: Calculate the energy required to raise the temperature of the ice to its melting point (0°C):
First, we need to calculate the heat absorbed by the ice using the formula:
Q = m * C * ΔT
Where:
Q is the heat absorbed (in Joules)
m is the mass of the ice (in grams)
C is the molar heat capacity of ice per gram (in J/(g⋅K))
ΔT is the change in temperature (in K)
Given:
m = 5.55 g
C = 36.6 J/(K⋅mol)
ΔT = (0 °C) - (-13.5 °C) = 13.5 K
To convert the molar heat capacity from J/(K⋅mol) to J/(g⋅K), divide it by the molar mass of ice:
Molar mass of H2O = 18.015 g/mol
C = 36.6 J/(K⋅mol) / 18.015 g/mol ≈ 2.030 J/(g⋅K)
Now, calculate the heat absorbed by the ice:
Q1 = m * C * ΔT
= 5.55 g * 2.030 J/(g⋅K) * 13.5 K
≈ 190.54 J
Since we need the answer in kilojoules (kJ), divide the value by 1000:
Q1 = 190.54 J / 1000
≈ 0.19054 kJ
Step 2: Calculate the energy needed to melt the ice and raise the temperature of the resulting water to the final temperature (26.5°C):
First, calculate the heat required to melt the ice:
Q2 = m * ΔH
Where:
Q2 is the heat absorbed (in Joules)
m is the mass of the ice (in grams)
ΔH is the heat of fusion of water (in J/g)
Given:
m = 5.55 g
ΔH = 6.01 kJ/mol (but we need it in J/g)
To convert the heat of fusion from kJ/mol to J/g, divide it by the molar mass of water (18.015 g/mol):
ΔH = 6.01 kJ/mol / 18.015 g/mol
≈ 0.333 J/g
Now, calculate the heat required to melt the ice:
Q2 = m * ΔH
= 5.55 g * 0.333 J/g
≈ 1.8472 J
Again, convert the value to kilojoules (kJ):
Q2 = 1.8472 J / 1000
≈ 0.0018472 kJ
Next, calculate the energy required to raise the temperature of the resulting liquid water from 0°C to 26.5°C:
Q3 = m * C * ΔT
= 5.55 g * 75.3 J/(g⋅K) * 26.5 K
≈ 10095 J
Again, convert the value to kilojoules (kJ):
Q3 = 10095 J / 1000
≈ 10.095 kJ
Finally, add up the three calculated values to get the total energy needed:
Total energy = Q1 + Q2 + Q3
= 0.19054 kJ + 0.0018472 kJ + 10.095 kJ
≈ 10.287 kJ
Therefore, approximately 10.287 kilojoules of energy are needed to heat 5.55 grams of ice from -13.5°C to 26.5°C.