19. Determine the final state and its temperature when 150.0 kJ of heat are added to 50.0 g of water at 20 ºC. (Specific heat of water = 4.184 J/g ºC; Specific heat of steam = 1.99 J/g • ºC; ∆Hfus (H2O) = 6.01 kJ/mol; ∆Hvap (H2O) = 40.79 kJ/mol). (5 points)

Question

06-07-2019
Physics

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19. Determine the final state and its temperature when 150.0 kJ of heat are added to 50.0 g of water at 20 ºC. (Specific heat of water = 4.184 J/g ºC; Specific heat of steam = 1.99 J/g • ºC; ∆Hfus (H2O) = 6.01 kJ/mol; ∆Hvap (H2O) = 40.79 kJ/mol). (5 points)

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hdawg7797 hdawg7797 · Answer 1 · 2019-07-06T23:06:49+02:00

The amount of energy to reach the boiling point is [tex]50*80*4.184 J=16,736J[/tex]. To pass the boiling point, [tex]40.79*\frac{50}{18.02}kJ=113,180J[/tex] are necessary (18.02 is the molar mass of water). This means that [tex]150kJ-113.180kJ-16.736kJ=20,084J[/tex] are left. This allows the steam to heat another [tex]\frac{20,084}{50*1.99}=201.8^{\circ}C[/tex]. Therefore, it ends as steam at temperature [tex]100^{\circ}C+201.8^{\circ}C=301.8^{\circ}C[/tex]