We calculated that 30.0 grams of ice initially at -10.0 Celsius and it requires 23.1 kj of energy to bring it all the way to a boil. How much additional energy does it take to convert all of the water to steam at 100 degree Celsius? How does this compare to what we calculated before?
It will take mass H2O x heat vaporization to turn the 30g into steam at 100 C.
To calculate the additional energy required to convert all of the water to steam at 100 degrees Celsius, we need to consider the heat required for the phase change from a liquid to a gas, known as the latent heat of vaporization.
The latent heat of vaporization for water is approximately 2260 kJ/kg. Since you have already calculated the mass of the ice (30.0 grams), we need to convert this to kilograms.
1 kg = 1000 grams
So, 30.0 grams = 30.0/1000 = 0.03 kg.
Next, we multiply the mass by the latent heat of vaporization to find the additional energy required:
Additional energy = mass * latent heat of vaporization
= 0.03 kg * 2260 kJ/kg
= 67.8 kJ
Therefore, it takes an additional 67.8 kJ of energy to convert all of the water to steam.
To compare this to the previous calculation, we calculated that it required 23.1 kJ of energy to bring the ice from -10 degrees Celsius to the boiling point. Comparing the two values, we can see that it takes significantly more energy to convert the water to steam compared to heating it from -10 degrees Celsius to the boiling point.