The boiling point of water is 1000C How much energy in kJ is

The boiling point of water is 100.0°C. How much energy (in kJ) is needed to heat 6.50 g of water to 122.7°C if the water initially was at 34°C? Liquid water: CP = 75.3 J/(mol · °C) Gas water: CP = 33.6 J/(mol · °C) Hvap = 40.67 kJ/mol MMwater = 18.02 g/mol

Solution

Answer :- 2.121 KJ

Notice that you must go from liquid to vapor,

Therefore the formula is,

                                                     q1 = m × C × T

                                        where,

q = the heat absorbed / given off

                                       m = the mass of the sample

c = the specific heat of the substance

                                    T = the change in temperature

                                              Final temperature – initial temperature

First we calculate the number of mole of water

6.50g ( 1 mol / 18.02 g ) = 0.3607 mol of H2O

The step describes your specific transition are,

Step 1 :- Going from water at 34 to water at 100

q1 = m × C(water) × T1

       = 0.3607 mol × 75.3 J/mol × (100-34)

       = 1792.66 J

Step 2 :- going from water at 100 to steam at 100

q2 = m × Hv

      = 0.3607mol × 40.67 kj/mol

     = 0.05321kJ

     = 53.21 J

Step 3 :- going from steam 100 to water at 122.7

q3 = m × C(steam) × T3

     = 0.3607mol × 33.6 J/mol × (122.7-100)

    = 275.11J

The total amount of energy required to go from ice at 34.0 to steam at 122.7 will be the sum of all the calculate energy

q(total) = q1 + q2 + q3

              = 1792.66 J + 53.21J + 275.11J

             =2120.98J

q(total) = 2.121 KJ