Question

At 1 atm, how much energy is required to heat 73.0 g of H2O(s) at –16.0 °C to H2O(g) at 167.0 °C? Helpful constants can be found here.

Answer 1

The ice at -16⁰C converted to 0⁰C

q1 = mcT

       = 73*2.1*(0-(-16)

          = 2452.8J

q2   = m*H fustion

         = 73*334   = 24382J

water converted to 0⁰C to 100⁰C

q3   = mcT

       = 73*4.184*(100-0)

         = 30543.2J

q4    = mH vap

          = 73*2260

           = 164980J

water at 100⁰C converted to 167⁰C

q5   = mcT

        = 73 *2.01*(167-100)

       = 9830.9J

Total energy is required q = q1 + q2 + q3 + q4 + q5

                                            = 2452.8 + 24382 + 30543.2 + 164980 + 9830.9

                                            = 232188   = 232.188KJ