Question

Water has the following thermodynamic values: ΔH°fus of H2O = 6.02 kJ/mol ΔH°vap of H2O = 40.7 kJ/mol heat capacity of solid H2O = 2.09 J/g°C heat capacity of liquid H2O = 4.18 J/g°C heat capacity of gaseous H2O = 1.97 J/g°C
How much energy (in kJ) is required to raise the temperature of 25.0 g of H2O from -129°C to 262°C?
Enter your answer in units of kJ to three significant figures.

Answer 1

Answer:

Qtotal = 90.004 kJ

Explanation:

To start resolving the problem we need to first convert the kJ/mol units from the thermodynamic values to J/g, so that we can work with the units of the heat capacity values. We know that the molar mass of water is 18.015 g/mol, so with this we do the respective conversion:

ΔH°fus of H2O = (6.02 kJ/mol) (1 mol/18.015g) (1000J/kJ) = 334.165 J/g

ΔH°vap of H2O = (40.7 kJ/mol) (1mol/18.015g) (1000J/kJ) = 2259.228 J/g

Now we need to find out the heat energy required to rise the temperature (specific heat capacity) and the energy required for each change of phase (specific latent heat), and add everything up. For this we will require the specific heat capacity and latent heat equations:

Q = mCΔT ; where m = mass, C = Hear capacity, ΔT = change of temperature

Q = mL ; where m = mass, L = specific latent heat

First change of phase (solid to liquid - fusion)

Q1 = (25g) (2.09 J/g°C) (0°C - (-129°C) = 6740.25 J

Q2 = (25g) (334.165 J/g) = 8354.125 J

Second change of phase (liquid to gas - vaporization)

Q3 = (25g) (4.18 J/g°C) (100°C - 0°C = 10450 J

Q4 = (25g) (2259.228 J/g) = 56480.7 J

Rise of temperature of the gaseous water

Q5 = (25g) (1.97 J/g°C) (262°C - 100°C = 7978.5 J

Finally we add everything up:

Qtotal = Q1 + Q2 + Q3 + Q4 + Q5 = 6740.25 J + 8354.125 J + 10450 J + 56480.7 J + 7978.5 J = 90003.575 J = 90.004 kJ