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

1. Jan 25, 2010

### 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.

2. Jan 26, 2010

### 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.