In an experiment, 10.0 grams of ice that is in at -10.0 degree C is added to 200

Question

In an experiment, 10.0 grams of ice that is in at -10.0 degree C is added to 200 grams of distilled water in a calorimeter. Before the ice was added the temperature of the distilled water was 22.0 degree C. After all the ice had melted, the temperature of the distilled water was 17.0 degree C. a. Using these results and the specific heat values for water and ice listed in the discussion, calculate Delta H_fus for ice. b. Now, given that the Delta H_fus for water is 6.02 kJ/mol, calculate the percent error for this experiment.

Explanation / Answer

Q = m c T
Q = quantity of heat in joules (J)
m = mass of the substance acting as the environment in
grams (g)
c = specific heat capacity (4.19 for H2O) in J/(g oC)

c = specific heat capacity of ice 2.06 J/(g oC)
T = change in temperature = Tfinal - Tinitial in oC

heat gained by ice + heat to melt ice + heat to raise water temp = heat lost by water

(22.00 g) (10 °C) (2.06 J/g °C) + (22.00 g / 18.0 g/mol) (x J/mol) + (22.00 g) ( 17 °C) (4.184 J/g °C) = (200.00 g) (4°C) (4.184 J/g °C)

453.2 + 1.22 x + 1564.816 = 3347.2

1.22x = 1329.184

x = 1089.49 joules

Question b

Percentage error = 100 - (1089.49 x 100 /6020) = 81.9 %

Related Questions

Navigate

• Browse (All)
• Browse I
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.