Determining Heat of Fusion for Ice: Calorimetry experiment

[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 transferred 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 affect 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.

 

[Steve4Physics]

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 specic heat of water
cA is the specic heat of the (aluminum) reservoir
q is the heat released by the thermometer which is equal to:

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

At the time of answering, this question is 12+ years old. But the following comments may be useful to anyone reading this thread.

The question has a number of faults, the main one being the lack of a description of the experiment. For example, it is unclear why the specific heat capacity and density of glass are given (these could relate to the stirrer and/or thermometer but their masses are not defined). So a small amount of guesswork is needed.

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.

The heat transfer from the themrometer is not included in the given equation - it is assumed to be negligible.

But since the thermometer is (probably) always initially warmer than the ice-water mixture, the thermometer will add some energy and the 'negligible' assumption may not be correct. This would introduce a systematic; error because an unknown amount of energy is always being added. The student should be able to determine its affect on the final answer, by imagining another term in the equation and noting its effect.

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 transferred 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.

Same comment as for the thermometer (above).

Also, the act of stirring adds (never remove) energy, so it introduces a systematic (not random) error. However the error is negligible for this sort of experiment.

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.

Yes. We assume that the only significant heat-transfer will be between the immersed part of the thermometer and the liquid as they are in direct contact.

4.) How does the amount of ice used affect 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.

The man point is that all the ice wouldn't melt.

Also note, a longer time would increase systematic error because a large source of error is the transfer of heat to the system from the lab' which is not a random direction.

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

A large amount of water would result in only a small temperature change for the water. The fractional uncertainty for the termpeature change would increase; this increases the uncertainty in the final result.

Edit - minor changes/typo's.