Question

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)

Answer 1

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