Need help with this practice problem. I got the values for the gas to liquid and liquid to solid. Don't know how to do the rest? Heat of Vap and fusion too? No idea!
What is the enthalpy change during the process in which 100.0 g of
water at 50.0 C is cooled to ice at 30.0 C?
The specific heats of ice, liquid water, and steam are 2.03 J/g-K, 4.18 J/g-K, and 1.84 J/g-K,
respectively. For H2O, Hfus = 6.01 kJ/mol and Hvap = 40.67 kJ/mol.
Please show the work. I know the answer is 60.4kJ
Thank you!
You have the phase changes. What you need is heat lost/gained within a phase. For both liquid and solid phase use this.
q = mass liquid H2O x specific heat H2O x (Tfinal-Tinitial)
q = mass ice x specific heat ice x (Tfinal-Tinitial)
By the way, ice at 30 C doesn't make sense. Neither does vap at 50 (it's a liquid there and not a gas).
To find the enthalpy change during the process from water at 50.0 °C to ice at -30.0 °C, we need to consider the different steps involved and calculate the energy for each step.
First, let's break down the process into three steps:
1. Cooling water from 50.0 °C to 0.0 °C (Liquid water)
2. Freezing water at 0.0 °C to ice at 0.0 °C (Solid water)
3. Cooling ice from 0.0 °C to -30.0 °C (Ice)
Step 1: Cooling water from 50.0 °C to 0.0 °C:
To find the energy required to cool the water, we use the formula:
q = m * ΔT * Cs
where q is the heat energy, m is the mass, ΔT is the change in temperature, and Cs is the specific heat.
For this step, we have:
m = 100.0 g (given)
ΔT = 0.0 °C - (-50.0 °C) = 50.0 °C
Cs (liquid water) = 4.18 J/g-K (given)
q1 = m * ΔT * Cs (Step 1)
q1 = 100.0 g * 50.0 °C * 4.18 J/g-K
Step 2: Freezing water at 0.0 °C to ice at 0.0 °C:
To find the energy required for this phase change, we use the formula:
q = m * ΔHfus
where q is the heat energy, m is the mass, and ΔHfus is the enthalpy of fusion.
For this step, we have:
m = 100.0 g (given)
ΔHfus = 6.01 kJ/mol (given)
First, we need to convert the mass to moles:
moles = m / molar mass of water
where molar mass of water = 18.02 g/mol
moles = 100.0 g / 18.02 g/mol
Then, we can calculate the energy:
q2 = moles * ΔHfus
q2 = (100.0 g / 18.02 g/mol) * 6.01 kJ/mol
Step 3: Cooling ice from 0.0 °C to -30.0 °C:
Similar to Step 1, we can use the formula:
q = m * ΔT * Cs
where q is the heat energy, m is the mass, ΔT is the change in temperature, and Cs is the specific heat.
For this step, we have:
m = 100.0 g (given)
ΔT = -30.0 °C - 0.0 °C = -30.0 °C
Cs (ice) = 2.03 J/g-K (given)
q3 = m * ΔT * Cs (Step 3)
q3 = 100.0 g * (-30.0 °C) * 2.03 J/g-K
Now, to find the total enthalpy change, we add up the energies from each step:
Total enthalpy change = q1 + q2 + q3
Finally, rearranging units and converting kJ to J, we get the answer:
Total enthalpy change = (q1 + q2 + q3) in J
Total enthalpy change = (q1 + q2 + q3) / 1000 in kJ
Calculate each step in J, then add the results and divide by 1000 to get the answer in kJ.