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, $40.79*\frac{50}{18.02}kJ=113,180J$ 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 $\frac{20,084}{50*1.99}=201.8^{\circ}C$. Therefore, it ends as steam at temperature $100^{\circ}C+201.8^{\circ}C=301.8^{\circ}C$