Question

B. 10 points In the Calorimetry lab you determined an experimental value for the heat of fusion of ice. The density of water is exactly 1 g/mL ATater is the temperature change of the water after the ice cube has melted. Tr-T, ke ATal is the total temperature change of the calorimeter Tr T, cal Use the data in the table to determine an experimental value for AHsi Determine a percent error if the known value is 6.02 kJ/mole Show your work. No work no credit Trial 1 Mass of calorimeter (g) 12.422 Mass of calorimeter + water 112.422 Mass of calorimter + water + ice 117.510 Mass of water Mass of ice (g) Moles of ice (n) Teal ("C) T ise (C) Tr (C) Cort (JC) 22.8 1.0 17.5 25.0 ATal C) ATwater (C) AT ice water CC) S water (J/gC) Aion (J/mol) Percent error

Answer 1

mass of calorimeter = 12.422 g

Mass of water = 112.422 - 12.422 = 100 g

Mass of ice = 117.51 - 112.422 = 5.088 g

Moles of ice (n) = 5.088 g/(18 g/mol) = 0.2827 mol

$\Delta$T = Tf - Ti

Where 'Ti' and 'Tf' are the corresponding initial and final temperatures

$\Delta$T_cal = 17.5 ^oC - 22.8 ^oC = -5.3 ^oC

$\Delta$T_water = 17.5 ^oC - 1 ^oC = 16.5 ^oC

The heat required for ice = (m.Cp.$\Delta$T)_ice

Where 'm' is the mass of ice

Cp is the specific heat capacity of ice = 4.18 J/g.^oC

i.e. The heat required for ice (q) = 5.088 g * 4.18 J/g.^oC * 16.5 ^oC = 350.91936 J

Now, $\Delta$H_fusion = q/n = 350.91936 J/0.2827 mol = 1241.46 J/mol or 1.24 kJ/mol

Therefore, percent error = {(6.02 - 1.24)/6.02}*100 = 79.4%