Question

A student measures the following data regarding the heat of fusion of ice: 24.6 g ice at 0.0°C is placed into the calorimeter which contains 100.0 g water at 22.6°C. The final temperature comes to 3.0°C. The calorimeter has a heat capacity of 15.7 J/°C.

(a) Calculate the experimental ΔH_fus of ice in kJ/mol.

Answer 1

solution::

no. of mole of ice = mass / molecular mass

= 24.6 / 18 = 1.366 mole

let deltaH_fusion of ice is xkJ / mol

hear neede to melt 24.6 g ice = 1.36 * x * 1000 J

hea needed to raise temprature of 24.6 g of water from 9 degre to 3 degree

= 24.6 * 4.2 * (3 - 0) = 309.96 J (use q = m* c_p * deltaT)

where q = hea, m = mass, c_p = specific heat, delta T = change in temeprature)

heat relased by 100 g of water to frop he temprarure from 22.6 degree to 3 degree is

= 100*4.2 *(22.6 - 3) = 8232 J

heat relesedby calorimeter

= 15.7 *(22.6 - 3) = 307.72 J

so, 1360 * x + 309.96 J = 8232 J + 307.72

x = 6.05

therefore

deltaH_fusion of ice is 6.05kJ / mol

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