Heat of fusion of water ice and dry ice (CO2) - need some insight

[Dumile]

I have decided to do away with the template; all calculations have been done, I'm just having trouble interpreting the results.

An experiment was performed using a calorimeter and warm water. The data collected gave the following values for heat of fusion for ice, and heat of sublimation of dry ice:

a) 17.06g of ice melted in 30.9°C water with mass 77.29g

Final water temp = 13.3°C.
ΔT = T(final) - T(initial) = 30.9 - 13.3 = -17.6°C

total energy absorbed(lost by water):
Q=4.18j/g°C x 77.29g x -17.6°C = 5686J.

Heat of fusion calculated:
L = Q/m = 5686J/17.06g = 333.3J/g <-- this is close to accepted value of 334J/g. So far so good.

b) 11.85g of ice melted in 13.3°C water with mass 94.35g.

Final water temp = 4.5°C. ΔT = -8.8°C

Total energy absorbed(lost by water) = 3471J.

Heat of fusion calculated = 292.9J/g <-- this is much lower than the accepted value. Why is this? Other than experimental error, is there a reason for this? I thought heat of fusion was constant between 0°C and boiling point.

c) 8.5g of dry ice (solid CO2) melted in 30.8°C water with mass 82.35g

final water temp = 15°C. ΔT = -15.8°C

Total energy absorbed(lost by water) = 5439J.

Heat of sublimation calculated = 663.3J/g. <-- accepted value is 571J/g. Why is the value higher? Is it because the evaporation process undergone by the dry ice cools the water to a greater degree, thus driving up the value of absorbed energy, which increases the value for latent heat?

I would like to understand what is happening here. Any insight would be great.

ps. Apologies if this thread is hard to follow - I'm new to all of this.

 

[kuruman]

Your analysis is based on the assumption that no heat escapes to the environment. Realistically, this is not so. How did you take into account the sure fact that heat also needs to be removed from the walls before the contents reach equilibrium? Most calorimetry experiments that I am familiar with involve determination of the "effective mass" of the calorimeter before one starts one's measurements.

 

[kerimek]

Thank you for providing the data from your experiment. It appears that you have correctly calculated the heat of fusion for water ice in the first part of your experiment, as it is close to the accepted value. However, in the second part of your experiment, the calculated heat of fusion is significantly lower than the accepted value. This could be due to experimental error, as you mentioned, but there could also be other factors at play.

One possible explanation for the lower heat of fusion in the second part of the experiment could be the presence of impurities in the ice. If the ice used in the second part of the experiment was not pure, it could have a lower heat of fusion due to the impurities interfering with the melting process. Additionally, the experimental conditions may have been slightly different, leading to a lower heat of fusion.

In the third part of your experiment, you have correctly calculated the heat of sublimation for dry ice. The higher value compared to the accepted value could be due to the fact that the evaporation process of dry ice is exothermic, meaning it releases heat into the surrounding environment. This could have contributed to a greater cooling effect on the water, resulting in a higher value for the heat of sublimation.

Overall, it is important to carefully consider all factors that could affect your results, such as experimental conditions and the purity of your materials. It is also helpful to compare your results to accepted values and to consider any potential sources of error. I hope this helps provide some insight into your experiment. Keep up the good work in your scientific pursuits!