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2) How much heat in kJ is
required to boil 50.0 grams of
water?

User Urlreader
by
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1 Answer

5 votes

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.

User JoeAC
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