# 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 59.4 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.

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1. Watch the units. You omitted the specific heat of liquid water. It is 4.184 J/g*C. Also you should convert 59.4 g H2O to mols H2O. mols = g/molar mass = ? and use the appropriate value for mass H2O (either mols or grams) with the specific heat (either J/g or kJ/mol).
q1 = heat to move T of ice from -50 to 0 C.
q2 = heat to melt ice
q3 = heat to move T liquid water from 0 C to 100 C
q4 = heat to boil & vaporize all of the water
q5 = heat to move T of H2O vapor from 100 C to 200 C.
q total = q1 + q2 + q3 + q4 + q5
q1 =mass ice(grams) x specific heat ice (J/g) x (Tfinal-Tnitial) = q in J.
q2 = mass ice (mol) x sp.h. (kJ/mol) = q in kJ. Convert to J.
q3 = mass water (g) x sp.h. (J/g) x (Tfinal-Tinitial) = q in J.
q4 = mass water(mols) x heat vaporization (kJ/mol) = q in kJ. Convert to J.
q5 = mass water (g) x sp.h. steam (J/g) x (Tfinal-Tinitial) = q in J.

Post your complete work if you get stuck.

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DrBob222
2. oops. q2 should be
q2 = mass ice (mol) x heat fusion (kJ/mol) = q in kJ. Convert to J.

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DrBob222

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