Question

In an experiment, 23.0 g of ice at –25°C is converted into steam with a temperature of 124°C. How much heat is required for the process? The following data may be of use:
vap = 2260 J/g; Hfus = 334 J/g; Cp(ice) = 2.06 J/g • °C; Cp(water) = 4.18 J/g • °C;
Cp(steam) = 1.99 J/g • °C

Answer 1

q1 = heat to move ice from -25 C to zero C.
q1 = mass ice x specific heat ice x delta T.

q2 = heat to melt ice (change H2O solid to H2O liquid at zero degrees.)
q2 = mass ice x heat fusion.

q3 = heat to move liquid water from zero degrees C to 100 C.
q3 = mass x specific heat water x delta T.

q4 = heat to convert liquid water at 100 C to steam at 100 C.
q4 = mass x heat vaporization.

q5 = heat to move steam from 100 C to 124 C.
q5 = mass x specific heat steam x delta T.

Total q = q1 + q2 + q3 + q4 + q5.