How many grams of ice at -24.3∘C can be completely converted to liquid at 23.9∘C if the available heat for this process is 4.82×103kJ ?
For ice, use a specific heat of 2.01J/(g⋅∘C) and ΔHfus=6.01kJ/mol.
To find the number of grams of ice that can be completely converted to liquid, we need to calculate the heat required to raise the temperature of the ice from -24.3∘C to 0∘C, then calculate the heat required to melt the ice, and finally calculate the heat required to raise the temperature of the resulting liquid water from 0∘C to 23.9∘C.
Step 1: Calculate the heat required to raise the temperature of the ice from -24.3∘C to 0∘C.
Q1 = mass × specific heat × ΔT1
Q1 = mass × 2.01 J/(g⋅∘C) × (0∘C - (-24.3∘C))
Q1 = mass × 2.01 J/(g⋅∘C) × 24.3∘C
Step 2: Calculate the heat required to melt the ice.
Q2 = mass × ΔHfus
Q2 = mass × 6.01 kJ/mol
Step 3: Calculate the heat required to raise the temperature of the resulting liquid water from 0∘C to 23.9∘C.
Q3 = mass × specific heat × ΔT3
Q3 = mass × 4.18 J/(g⋅∘C) × (23.9∘C - 0∘C)
Q3 = mass × 4.18 J/(g⋅∘C) × 23.9∘C
Step 4: Add up the heat required for each step.
Q_total = Q1 + Q2 + Q3
Now we can substitute the values and solve for mass:
4.82 × 10^3 kJ = mass × 2.01 J/(g⋅∘C) × 24.3∘C + mass × 6.01 kJ/mol + mass × 4.18 J/(g⋅∘C) × 23.9∘C
First, we need to convert the units of specific heat and heat of fusion to kJ/(g⋅∘C):
2.01 J/(g⋅∘C) = 0.00201 kJ/(g⋅∘C)
6.01 kJ/mol = 0.00601 kJ/(g⋅∘C)
4.18 J/(g⋅∘C) = 0.00418 kJ/(g⋅∘C)
Now we can substitute the converted values and solve for mass:
4.82 × 10^3 kJ = mass × 0.00201 kJ/(g⋅∘C) × 24.3∘C + mass × 0.00601 kJ/(g⋅∘C) + mass × 0.00418 kJ/(g⋅∘C) × 23.9∘C
Simplifying the equation:
4.82 × 10^3 = mass × 0.0489303 + mass × 0.00601 + mass × 0.099802
4.82 × 10^3 = mass × (0.0489303 + 0.00601 + 0.099802)
4.82 × 10^3 = mass × 0.1547423
mass = (4.82 × 10^3) / (0.1547423)
mass ≈ 31189.44 g
Therefore, approximately 31189 grams of ice at -24.3∘C can be completely converted to liquid at 23.9∘C if the available heat for this process is 4.82×103kJ.
To determine the number of grams of ice that can be completely converted to liquid, we need to calculate the energy required to melt the ice and compare it with the available heat.
The energy required to melt the ice (ΔH) can be calculated using the equation:
ΔH = n * ΔHfus
where ΔH is the energy required, n is the number of moles, and ΔHfus is the molar enthalpy of fusion.
First, we need to find the number of moles of ice by using the molar mass of water (18.015 g/mol):
n = mass / molar mass
To find the mass of ice, we use the equation:
mass = q / ΔT
where q is the available heat and ΔT is the change in temperature.
To calculate the change in temperature (ΔT), we subtract the initial temperature of the ice (-24.3∘C) from the final temperature of the liquid water (23.9∘C):
ΔT = Tf - Ti
Substituting these values into the equations, we can find the mass of ice.
q needed to raise T of ice at -24.3 to zero is q1 = mass ice x specific heat ice x (Tfinal-Tinitial). We don't know mass ice yet but this equation would look like this.
q1 = mass ice x 2.01 x [0-(-24.3)] = mass ice x 48.84 = 48.84m
q2 = heat needed to melt ice at zero to liquid water at zero.
q2 = mass ice x heat fusion. The heat fusion is given in the probem as 6.01 kJ/mol and that needs to be converted to J/g. That number is 334 J/g*C. We don't know mass ice yet. The equation would look like this.
q2 = mass ice x 334 J/g = 334m.
q3 = head need to raise T of liquid water at zero C to 23.9 is
q3 = mass H2O x specific heat H2O x (Tfinal-Tinitial). We don't know mass liquid water but it will be the same as mass ice. Substitute as I've done for q1 and q2 to find q3 in terms of m.
Add q1 + q2 + q3 = total q needed in terms of mass ice and set that = 4820000 (the available heat) and solve for m = mass ice in grams.
Post your work if you get stuck.