Answer:
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
