I have decided to do away with the template; all calculations have been done, I'm just having trouble interpreting the results.(adsbygoogle = window.adsbygoogle || []).push({});

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.

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# Homework Help: Heat of fusion of water ice and dry ice (CO2) - need some insight

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