Question

Calculate the amount of energy (in kJ) necessary to convert 377 g of liquid water from 0°C to water vapor at 172°C. The molar heat of vaporization (Hvap) of water is 40.79 kJ/mol. The specific heat for water is 4.184 J/g · °C, and for steam is 1.99 J/g · °C. (Assume that the specific heat values do not change over the range of temperatures in the problem.)

Answer 1

Answer: The amount of energy required is 1066 kJ.

Step-by-step explanation:

Our 377 g of liquid water will pass through three distinct stages as it heats:

• Stage 1: Heating liquid water from 0°C to 100°C (the boiling point of water).
• Stage 2: Vaporizing the liquid water at 100°C into steam at 100°C.
• Stage 3: Heating steam from 100°C to 172°C.

We must track the energy required over each stage separately.

Stage 1:

We have 377 g of liquid water at 0°C and will heat it to 100°C to reach boiling point. We know that the specific heat for water is 4.184 J/g · °C.

Using the heat equation,
`q = mc\Delta T`, we have:

`q = (377 \text{ g})(4.184 \text{ J/(g}\cdot {}^\circ\text{C}))(100 \ {}^\circ\text{C} - 0 \ {}^\circ\text{C}) = 157736.8 \text{ J} = 1.577 * 10^5 \text{ J}`

Stage 2:

We must vaporize the liquid water at 100°C into steam at 100°C. We know that the molar heat of vaporization of water is 40.79 kJ/mol.

377 g H2O x 1 mol H2O / (18.016 g H2O) x 40.79 kJ / mol = 853.6 kJ

Stage 3:

Now, we must finish the task by heating steam from 100°C to 172°C. We know that the specific heat for steam is 1.99 J/g · °C.

Using the heat equation,
`q = mc\Delta T`, we have:

`q = (377 \text{ g})(1.99 \text{ J/(g}\cdot {}^\circ\text{C}))(172 \ {}^\circ\text{C} - 100 \ {}^\circ\text{C}) = 54016.56 \text{ J} = 5.402 * 10^4 \text{ J}`

Conclusion:

We now calculate the total amount of energy in kJ. We must convert our Stage 1 and Stage 3 calculations into kJ first. Then,

`157.7 \text{ kJ} + 853.6 \text{ kJ} + 54.02 \text{ kJ} = 1065.32 \text{ kJ} = 1066 \text{ kJ}`

after accounting for significant digits.