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Answer :
The heat change for vaporization of 6.00 kg of water at 100°C is 13490.584 kJ.
The heat or enthalpy for vaporization is the amount of heat energy which is required to bring change in the state of a substance, such as from a liquid into a vapor or gas. It is also known as enthalpy of vaporization. The unit of enthalpy for vaporization is given in joules (J) or calories (cal).
Given data :
mass of water = 6.00 kg
Hvap = 40.7 kJ/mol
To calculate the amount of heat -
we know,
the heat required for 1 mol of water to vaporize at 100°C is 40.7 kg
and
we know molar weight of water is 18.015 gram
So,
moles of 6.00 kg water is = 2600
18.1015
mole of water = 331.464 mol
Hence,
Heat change for vaporization is = 331.464 × 40.7
= 13490.584 kJ
To learn more about vaporization,
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