# Determining Heat of Fusion for Ice:Calorimetry experiment

1. Feb 7, 2010

### pann2310

Hi everyone,

I'm taking a first year physics course and our upcoming lab is an experiment to determine the Heat of fusion for ice. There are a few questions that I have.

Firstly the overall equation is given as Lfm + cm (T2 - 0) = cmw (T1 - T2) + cAmA (T1 - T2) + q
Where:

m is the mass of ice
mw is the mass of water
mA is the mass of the reservoir and stirrer (aluminum)
c is the speci c heat of water
cA is the speci c heat of the (aluminum) reservoir
q is the heat released by the thermometer which is equal to:

q = cgmg (T1 - T2)
cg is the speci c heat of glass 837J=(kg*K)
g is the density of glass 2.3g=cm^3
q = 1.93V (T1 -T2)

1.) What assumptions have been made in calculating the heat given off by the thermometer and will this introduce a random or systemic and if systemic will it make calculated value of Lf higher or lower.

I am uncertain as to the assumptions made in calculating the heat given off by the thermometer. I do believe this will introduce a systemic error and that it could either make Lf higher or lower depending on the actual heat given off.

2.) What assumptions have been made about the temperature of the stirrer. Will this introduce a random or systemic and if systemic will it make calculated value of Lf higher or lower.

I believe the assumption being made is that no energy is being transfered from the stirrer to the system. The action of stirring will add kinetic energy to the system and cause a random error (as the amount of kinetic energy is dependent on the stirrer and can be considered random). This error will make Lf higher than it should be.

3.) Why is it only the immersed volume of the thermometer that matters?

Because its a closed system only the elements within the system should transfer heat which is the immersed volume of the thermometer.

4.) How does the amount of ice used a ffect the results? What would happen if a large amount of ice were used with a small amount of water and vice versa?

If you use a large amount of ice, then the amount of energy in the system may not be enough to induce a phase change. Also if there is enough energy it would take a long time to melt the ice and would take a longer time to record the data. This would increase the probability of random errors occurring.

I believe a large amount of water and small amount of ice would make observing the phase change difficult to measure. Also I am uncertain how the initial temperatures of the system and the ice come into play with this question. As in a small amount of water would mean that the initial temperature would be low and vice versa.

Any help or clarification of these questions would be greatly appreciated.