The quantity of heat released is 63.09 kJ.
n = 27.9 g H₂O × (1 mol H₂O/18.02 g H₂O) = 1.548 mol H₂O
ΔH_cond = -ΔH_vap = -40.7 kJ·mol⁻¹
The negative sign shows that the water vapour is releasing energy as it condenses.
q = nΔH_cond = 1.548 mol × (-40.7 kJ·mol⁻¹) = -63.0 kJ
Answer:
63.09 kJ is the quantity of heat that is released when 27.9 g of water condenses.
Explanation:
[tex]H_2O(l)\rightarrow H_2O(g),\Delta H_{vap}=40.7 kJ/mol[/tex]
Latent heat of vaporization =[tex]\Delta H_{vap}=40.7 kJ/mol[/tex]
Amount of heat required to condense 1 mole of water = 40.7 kJ
[tex]H_2O(g)\rightarrow H_2O(l),\Delta H_{vap}=-40.7 kJ/mol[/tex]
Mass of water given = 27.9 g
Moles of water :
[tex]\frac{27.9 g}{18 g/mol}=1.55 mol[/tex]
Heat required to vaporize 1.55 moles of water:
[tex]1.55 moles\times (-40.7 kJ/mol)=-63.085 kJ\approx -63.09 kJ[/tex]
Negative sign indicates that energy will release.
