Question

Calculate the enthalpy of fusion (KJ/mole) for ice using the following lab data:
(HINT: Think about the heating curves for H20)

Mass of empty calorimeter + stir bar: 255.060 g
Mass of calorimeter, stir bar, and warm water: 309.602 g
Mass of calorimeter, stir bar, water, and melted ice: 312.170g
Mass of warm water in calorimeter: 54.543 g
Mass of ice added: 2.568 g
The initial temperature of water: 35.3 C
The final temperature of water/melted ice: 31.1 C
Specific heat of water: 4.184 J
Hfusion of water: -285

Answer 1

see clearly there is something wrong.I have checked my calculation twice..can't really find out the issue.

The one probable issue can be due to unavailability of specific heat of empty calorimeter and stir bar. Those two also come down to lower temperature just to maintain equilibrium.I have marked in red ink where that should be added.Once you add that u will get values close to actual one.

Another issue is that Hfusion of water..what is that actually? and what is the unit in which it is given?These clarifications are needed.

But I repeat concept is as follows..

Warm water, calorimeter and stir bar gives heat into the system and end up in a lower temperature. Ice collects that heat.uses part of it in melting with no change in temperature and rest of the heat to come in equilibrium with warm water,calorimeter and stir bar too.

Most importantly we are assuming the whole process is happening in a closed chamber where no heat can escape.(This is actually impossible in practical cases of laboratory)

Thank you! Cheers.