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Anna007 [38]
3 years ago
5

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)
Physics
1 answer:
8090 [49]3 years ago
6 0

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

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