Posted by dinah on .
At 1 atm, how much energy is required to heat 79.0 g of H2O(s) at –20.0 °C to H2O(g) at 121.0 °C?
figure the heat to take ice to OC
figure the heat to melt the ice at OC
figure the heat to warm it from 0C to 100C
figure the heat to vaporize the water at 100C
Figure the heat to heat the steam from 100C to 121C.
add them together.
remember, ice, water, steam all have different specific heat capacities c.
q1 = heat to raise T solid H2O from -20 to ice at zero C.
q1 = mass H2O x specific heat ice x (Tf-Ti) where Tf = 0 and Ti = -20
q2 = heat to melt ice.
q2 = mass ice x heat fusion.
q3 = heat to raise T of liquid H2O from zero C to 100 C.
q3 = mass H2O x specific heat liquid H2O x (Tfinal-Tinitial) where Tf = 100 and Ti = 0.
q4 = heat to change liquid H2O at 100 C to steam at 100 C.
q4 = mass H2O x heat vaporization.
q5 = heat to raise steam from 100 C to 121 C.
q5 = mass steam x specific heat steam x (Tfinal-Tinitial) where Tf is 121 C and Ti is 100 C.
Total q = q1 + q2 + q3 + q4 + q5