Question

Calculate the amount of energy (in kJ) necessary to convert 297 g of liquid water from 0°C to water vapor at 162°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

To convert 297 g of liquid water from 0°C to water vapor at 162°C, the total energy required is 796.02 kJ. This can be calculated by considering the energy required to heat the water from 0°C to 100°C and the energy required to vaporize the water at 100°C.

Step-by-step explanation:

To calculate the amount of energy necessary to convert 297 g of liquid water from 0°C to water vapor at 162°C, we need to consider two processes:

1. The energy required to heat the water from 0°C to 100°C.
2. The energy required to vaporize the water at 100°C.

For the first process, we will use the specific heat for water, which is 4.184 J/g · °C.

Since the temperature change is 100°C, the energy required is:

Q1 = m * C * ΔT

= 297 g * 4.184 J/g · °C * 100°C

= 124,051.76 J

For the second process, we will use the molar heat of vaporization (Hvap) of water, which is 40.79 kJ/mol. We need to convert the mass to moles using the molar mass of water, which is 18.015 g/mol:

n = m / M

= 297 g / 18.015 g/mol

= 16.483 mol

The energy required to vaporize the water is:

Q2 = n * Hvap

= 16.483 mol * 40.79 kJ/mol

= 671.97 kJ

The total energy required is the sum of Q1 and Q2:

Total energy = Q1 + Q2 = 124,051.76 J + 671,970 J

= 796,021.76 J

= 796.02 kJ