Question

At 1 atm, how much energy is required to heat 59.0 g of H2O(s) at –14.0 °C to H2O(g) at 153.0 °C?

Answer 1

mass of water = 59.0 g

moles of water = 59 / 18 = 3.28

-14.0 °C ----------> 0.0 °C ------------> 100.0 °C -------------> 153.0 °C

                 1           2          3             4              5

in this process 5 conversions are required.

Q1 = m Cp dT = 59 x 2.09 x (0 + 11) = 1356.41 J

Q2 = n x delta H fusion = 3.28 x 6.01 x 10^3 = 19712.8 J

Q3 = m Cp dT = 59 x 4.18 x (100 - 0) = 24662 J

Q4 = n x delta H vap = 3.28 x 40.7 x 10^3 = 133496 J

Q5 = m Cp dT = 59 x 2.01 x (153 - 100) = 6285.3 J

total heat = Q1 + Q2 + Q3 + Q4 + Q5

= 185512.5 J

= 186 kJ