# AP Chemistry

Given that the specific heat capacities of ice and steam are 2.06 J/g ·◦C and 2.03 J/g·◦C,respectively, calculate the total quantity of heat necessary to melt 19.2 g of ice at −50.0
◦C and heat it to steam at 200◦C. The molar heats of fusion and vaporization for water are
6.02 kJ/mol and 40.6 kJ/mol, respectively. Assume water freezes at 0◦C and boils at 100◦C.
Answer in units of J.

PLEASE DO NOT ROUND WHEN SHOWING STEPS

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1. A multistep problem.
q1 = heat needed to raise T ice from -50 to zero C.
q1 = mass ice x specific heat ice x (Tfinal-Tinitial)

q2 = heat needed to change solid ice at zero C to liquid water at zero C.
q2 = mass ice x heat fusion ice.

q3 = heat needed to raise T of liquid H2O from zero C to 100 C.
q3 - mass H2O x specific heat liquid H2O x (Tfinal-Tinitial)

q4 = heat needed to change liquid water at 100 C to steam at 100 C.
q4 = mass H2O x heat vaporization

q5 = heat needed to raise T of steam at 100 C to steam at 200 C.
q5 = mass steam x specific heat steam x (Tfinal-Tinitial)

Total Q = q1 + q2 + q3 + q4 + q5

IMPORTANT NOTE:
Steps 1,3,5 are "within a single phase" and the formula is exactly the same for each except the numbers are different because you are dealing in 1 with solid ice, in 3 with liquid water and in 3 with steam. Anytime you have no phase change, the formula will calculate q for you.

At the melting point AND the boiling point, you have a PHASE CHANGE. A phase change from solid ice to liquid water is step 2 and a phase change from liquid water to steam is step 4. Anytime you have a phase change, the formula is the same except at the melting point you use the heat of fusion and at the boiling point you use the heat of vaporization.
Those two formula will work ALL of the problem of this type.

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