Answer:
The heat required to boil water is called the heat of vaporization, and it is the amount of energy required to change the state of a substance from a liquid to a gas at a constant temperature. The heat of vaporization for water is approximately 40.7 kJ/mol or 2257 J/gram.
To calculate the heat required to boil 50.0 grams of water, you would multiply the heat of vaporization (2257 J/g) by the number of grams of water:
Heat (J) = 2257 J/g x 50 g
Heat (J) = 112,850 J
To convert Joules to kilojoules (kJ), divide by 1000.
Heat (kJ) = 112,850 J / 1000 = 112.85 kJ
Therefore, it takes 112.85 kJ of heat to boil 50.0 grams of water.