How much energy is required to convert 15.0 g of ice at - 106 °C to water vapor at 125 °C? Specific heats are 2.09 J/g K for both ice and water vapor, and 4.18 J/g K for liquid water. Heats of vaporization and fusion are 2.260 kJ/g and 0.335 kJ/g, respectively.

49.3 kJ of energy is required

Explanation:

An exercise of calorimetry at its best

First of all, convert the ice to water before melting.

Q = ice mass. C. ΔT

Q = 15 g. 2.09 J/g°C (0° - (-106°C)

15 g. 2.09 J/g°C. 106°C = 3323.1 J

Now we have to melt the ice, to change its state

Q = mass. latent heat of fusion

Q = 15 g. 0.335 kJ/g = 5.025 kJ. 1000 = 5025 J

After that, we have liquid water at 0° and the ice has melted completely. We have to release energy to make a temperature change, to 100° (vaporization)

Q = 15g. 4.18 J/g°C (100°C - 0°C)

Q = 6270 J

Water has been vaporizated so we have to calculate, the state change.

Q = mass. latent heat of vap

Q = 15 g. 2.260 kJ/g

Q = 33.9 kJ (.1000) = 33900 J

Finally we have to increase temperature from 100°C to 125°C

Q = 15 g. 2.09 J/g°C. (125°C - 100°C)

Q = 783.75 J

To know how much energy is required to conver 15 g of ice, to water vapor at 125°C, just sum all the heat released.

3323.1 J + 5025 J + 6270 J + 33900 J + 783.75 J = 49301.85 joules.

Notice I have to convert kJ to J in two calcules to make the sum.

49301.85 joules / 1000 = 49.3 kJ