Answer:
535.3 kJ heat was absorbed.
Step-by-step explanation:
The heat absorbed by the water can be calculated using the equation:
Q = mLf
where Q is the heat absorbed, m is the mass of the water, Lf is the heat of vaporization, and Lf is the amount of heat required to vaporize a substance at a constant temperature.
The heat of vaporization of water at 100°C is 40.7 kJ/mol. Since 1 mol of water has a mass of 18.02 g, the heat of vaporization per gram of water is 40.7 kJ/mol / 18.02 g/mol = 2.257 kJ/g.
So the heat absorbed by the 238.8 g of water can be calculated as:
Q = 238.8 g * 2.257 kJ/g = 535.3 kJ.
This is the amount of heat energy absorbed by the water to convert from a liquid at 17.5°C to a gas at 118.9°C.