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The heat of vaporization of water at 100°c is 40.66 kj/mol. Calculate the quantity of heat that is absorbed/released when 5.00 g of steam condenses to liquid water at 100°c.

Answer :

Answer : The quantity of heat released is -11.30 kJ

Explanation :

First we have to calculate the number of moles of water.

[tex]\text{Moles of water}=\frac{\text{Mass of water}}{\text{Molar mass of water}}[/tex]

Molar mass of water = 18 g/mole

[tex]\text{Moles of water}=\frac{5.00g}{18g/mole}=0.278mole[/tex]

Now we have to calculate the amount of heat released.

[tex]\Delta H=-\frac{q}{n}[/tex]

where,

[tex]\Delta H[/tex] = heat of vaporization = 40.66 kJ/mol

q = heat released = ?

n = number of moles of water = 0.278 mole

[tex]40.66kJ/mol=-\frac{q}{0.278mol}[/tex]

[tex]q=-11.30kJ[/tex]

In vaporization process, the amount of heat is absorbed but in the process of condensation the amount of heat is released.

Therefore, the quantity of heat released is -11.30 kJ

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