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Determine the amount of heat (in kj) required to vaporize 1.55 kg of water at its boiling point. for water, δhvap

Respuesta :

Heat required to vaporize 1 mol of water from water at 100C to steam at 100C = 40.7 kJ 
1 mol of water weighs =  18.015g
1.55 kg = 1550/18.015 mol = 86.03 mol 

Heat required to vaporize :
= 86.03 mol  x 40.7 kJ 

= 3501.421 kJ
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From the calculation and data provided, the heat of vaporization of water is 3.5 * 10^6 J.

What is heat of vaporization?

The heat of vaporization is the heat that is required to convert a given mass of water from liquid to gas.

Given that we have 1.55 kg of water and the latent heat of vaporization of water is  22.6 x 10^5 J/kg.

Thus;

Hvap = 1.55 kg *  22.6 x 10^5 J/kg = 3.5 * 10^6 J

Learn more about heat of vaporization:https://brainly.com/question/12625048

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