Question

At 1 atm, how much energy is required to heat 81.0 g of H2O(s) at –18.0 °C to H2O(g) at 137.0 °C?

Answer 1

mass of water = 81 g

moles of water = 81 / 18 = 4.5

-18.0 °C ----------> 0.0 °C ------------> 100.0 °C -------------> 137.0 °C

                 1           2          3             4              5

in this process 5 conversions are required.

Q1 = m Cp dT = 81 x 2.09 x (0 + 18) = 3047 J

Q2 = n x delta H fusion = 4.5 x 6.01 x 10^3 = 27045 J

Q3 = m Cp dT = 81 x 4.18 x (100 - 0) = 33858 J

Q4 = n x delta H vap = 4.5 x 40.7 x 10^3 = 183150 J

Q5 = m Cp dT = 81 x 2.01 x (137 - 100) = 6023.97 J

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

               = 3047 + 27045 + 33858 + 183150 + 6023.97

               = 253123.9 J

               = 253 kJ