Determining Heat of Fusion for Ice: Calorimetry experiment

In summary, the upcoming lab is an experiment to determine the Heat of fusion for ice, and the overall equation given includes various variables such as mass, specific heat, and heat released by the thermometer. The assumptions made in calculating the heat given off by the thermometer and the temperature of the stirrer may introduce systematic errors and affect the calculated value for Lf. The immersed volume of the thermometer is the only important factor in the experiment as it is a closed system. Using a large amount of ice or water may affect the results due to the amount of energy needed for a phase change and the difficulty in measuring it. Clarification and assistance with these questions would be appreciated.
  • #1
pann2310
1
0
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 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.
 
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  • #2
pann2310 said:
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.

pann2310 said:
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.

pann2310 said:
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.

pann2310 said:
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.

pann2310 said:
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.

pann2310 said:
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.
 
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1. What is the purpose of determining the heat of fusion for ice through a calorimetry experiment?

The purpose of this experiment is to measure the amount of heat energy required to melt a certain amount of ice. This is important in understanding the physical properties of ice and can also be used to calculate the specific heat capacity of water.

2. How is the heat of fusion for ice determined through a calorimetry experiment?

The heat of fusion for ice is determined by measuring the change in temperature of a known mass of ice as it melts in a calorimeter. The amount of heat energy absorbed by the ice is then calculated using the equation Q = mCΔT, where Q is the heat energy, m is the mass of the ice, C is the specific heat capacity of water, and ΔT is the change in temperature.

3. What equipment is needed for a calorimetry experiment to determine the heat of fusion for ice?

The equipment needed for this experiment includes a calorimeter, a thermometer, a known mass of ice, and a heat source such as a Bunsen burner or hot plate. The calorimeter should be insulated to minimize heat loss during the experiment.

4. What are the potential sources of error in a calorimetry experiment to determine the heat of fusion for ice?

Potential sources of error in this experiment include heat loss to the surroundings, incomplete melting of the ice, and not accounting for the heat capacity of the calorimeter. It is important to minimize these errors by using proper insulation and ensuring the ice is completely melted.

5. How can the results of a calorimetry experiment to determine the heat of fusion for ice be used in real-world applications?

The heat of fusion for ice is an important value in industries such as food preservation and refrigeration. It is also used in weather forecasting and understanding the Earth's climate. This experiment can also be used to calculate the specific heat capacity of water, which has practical applications in heating and cooling systems.

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