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!
To solve this problem, we need to consider the different steps involved in the process of cooling water from 50.0 °C to -30.0 °C. The steps are:
1. Cooling the water from 50.0 °C to 0 °C
2. Freezing the water at 0 °C to ice at 0 °C
3. Cooling the ice from 0 °C to -30.0 °C
Let's calculate the enthalpy change for each step and add them up to find the total enthalpy change.
Step 1: Cooling the water from 50.0 °C to 0 °C
To calculate the heat lost during cooling, we can use the formula:
Q = m × C × ΔT
where Q is the heat lost or gained, m is the mass, C is the specific heat, and ΔT is the change in temperature.
Given:
Mass of water (m) = 100.0 g
Specific heat of liquid water (C) = 4.18 J/g-K
Change in temperature (ΔT) = 0 °C - 50.0 °C = -50.0 °C
Q1 = 100.0 g × 4.18 J/g-K × (-50.0 °C)
Q1 = -20900 J
Note: The negative sign indicates heat lost by the water.
Step 2: Freezing the water at 0 °C to ice at 0 °C
The heat released during the freezing process is known as the heat of fusion (ΔHfus).
To calculate the heat released when freezing, we can use the formula:
Q = ΔHfus × moles
First, we need to calculate the moles of water:
1 mole of water = 18.02 g
moles = mass (g) / molar mass
moles = 100.0 g / 18.02 g/mol
moles ≈ 5.55 mol (approx.)
Given:
ΔHfus = 6.01 kJ/mol
Q2 = 6.01 kJ/mol × 5.55 mol
Q2 = 33.34 kJ
Step 3: Cooling the ice from 0 °C to -30.0 °C
Using the same formula as in Step 1:
Given:
Mass of ice (m) = 100.0 g
Specific heat of ice (C) = 2.03 J/g-K
Change in temperature (ΔT) = -30.0 °C - 0 °C
ΔT = -30.0 °C
Q3 = 100.0 g × 2.03 J/g-K × (-30.0 °C)
Q3 = -6060 J
Now, let's calculate the total enthalpy change by summing up the individual enthalpy changes:
Total enthalpy change = Q1 + Q2 + Q3
Total enthalpy change = -20900 J + 33.34 kJ - 6060 J
Total enthalpy change ≈ 60.4 kJ
So, the enthalpy change during the process is approximately 60.4 kJ, which matches the given answer.