Heat of Fusion    

Ben Conway: ben150jc@aol.comRocky Goetz: rgoetz10@hotmail.com

Jesse Bennett: jesse-bennett@hotmail.com 

Performed on 2-21-12Due 2-28-12

         

AbstractThis experiment dealt with the theory concerning the latent heat of fusion for water. The

objective of this experiment was to determine the latent heat of fusion of water by mixing ice and

warm water. The result of this experiment confirmed the theory by giving a latent heat of 80.4

cal/g with a small percent error of 0.86% from the accepted value of 80.0 cal/g.

  

Theory:Heat of fusion is defined by:

The quantity of heat absorbed or released by a substance undergoing a change of state. Such as

ice changing to water, or water to steam, at constant temperature and pressure. Also called heat

of transformation.

In the mixture of ice and water it is assumed that heat is conserved. To determine the latent heat

of fusion for water we examine the “heat lost=heat gained” equation:𝑀,𝐿-𝑓.+𝑀,𝑠-𝑤.,,𝑡-2.−0.=(,𝑚-𝑠.,𝑆-𝑤.+,𝑚-1.,𝑠-1.+,𝑚-2.,𝑠-2.)(,-1.−,-2.)

Where M is the mass of ice, in grams; ,-𝑓. is the latent heat of fusion of ice in calories per gram,

,𝑡-2. is the equilibrium temperature in ℃, ,𝑡-1. is the initial temperature of the water and

colorimeter, ,𝑚-𝑤. is the mass of warm water and ,𝑠-𝑤. is its specific heat (1 cal/℃-gm), ,𝑚-

1.,𝑠-1.+,𝑚-2.,𝑠-2. is the water equivalent of the calorimeter and stirrer, ,𝑚-1. is the mass and

,𝑠-1. is the specific heat of the calorimeter, ,𝑚-2. is the mass and ,𝑠-2. is the specific heat of the

stirrer.

Equipment:

1. Calorimeter

2. Stirrer

3. Thermometer

4. Scale

                                                                   Procedure:1)    The water in the colorimeter temperature was measured. Then several ice cubes were added.

Stirring the water take the temperature when it is 10 to 15℃ below room temperature. Then

obtain the mass of the ice by measuring the weight again.

The procedure was repeated two more times.                                                                     

DataTable 1

Data Table Trial 1 Trail 2 Trial 3Mass of calorimeter = m1 (g) 116.6 116.6 116.6Mass of water and calorimeter (g)

277.8 275.0 274.3

Mass of water = mw (g) 120.8 118.0 117.3Mass of resulting solution (g) 146.3 141.3 143.6Mass of ice = M (g) 26.6 23.9 27.1Mass of stirrer = m2 (g) 3.6 3.6 3.6Initial temperature = t1 (*C) 22 20.8 25.7final temperature = t2 (*C) 6.4 6.2 8.4Latent heat of fusion = Lf (cal/g)

 79.6 78.6  83.0 

Table 1 is the data recorded during each trial. 

  

AnalysisThe formula given in the theory section is what was used to determine the latent heat of fusion for water in this experiment:𝑀,𝐿-𝑓.+𝑀,𝑠-𝑤.,,𝑡-2.−0.=(,𝑚-𝑠.,𝑆-𝑤.+,𝑚-1.,𝑠-1.+,𝑚-2.,𝑠-2.)(,-1.−,-2.)When you solve for the latent heat the formula becomes:

,𝐿-𝑓.=,(𝑚-𝑠.,𝑆-𝑤.+,𝑚-1.,𝑠-1.+,𝑚-2.,𝑠-2.),,,𝑡-1.−,𝑡-2..-𝑀.−,,𝑠-𝑤.,,𝑡-2..-𝑀.Sample calculation:

,𝐿-𝑓.=,,96.0𝑔∗,1𝑐𝑎𝑙-℃𝑔𝑚.+115.6𝑔∗,0.2154𝑐𝑎𝑙-℃𝑔𝑚.+38.9𝑔∗,0.2154𝑐𝑎𝑙-℃𝑔𝑚..,25.7−12.3℃.−17.3𝑔,,1𝑐𝑎𝑙-℃𝑔𝑚..(12.3℃)-17.3𝑔. ,-𝑓.=87.8 cal/g

  

Results  

After repeating the experiment 3 times the average of the 3 results were found by adding them together and dividing by 3:

,𝑣𝑔𝑒𝑟𝑎𝑔𝑒 𝐿-𝑓.=80.4cal/gThe actual accepted value of the latent heat of fusion for water is 79.72cal/g. So it is helpful to find the percent error of this experiment:𝑃𝑒𝑟𝑐𝑒𝑛𝑡 𝐸𝑟𝑟𝑜𝑟=0.86%This is a relatively low percent error so this experiment confirms the theory of the latent heat of

fusion of water.  

Discussion 

The percent error for this experiment was very low, suggesting the execution was well and the experiment was well understood. One way to make the experiment even better would be to make the container 100% sealed and insulated so that there is no chance of heat escape to the surroundings.Questions:

1.     The estimates of precision for the temperature fall of the water would be around + or - 0.2%. And for mass of ice around + or – 0.2%. So the estimate for ,-𝑓. would be about 0.4%.

2.     The percent discrepancy was calculated above to be 0.86%3. σ = 1.88

Lave±σ = 80.4±1.88 cal/g4.     a.) If the ice was taken directly from the freezer then it probably would seem to have a lower

mass because of the air trapped inside it which would make the latent heat of fusion calculation too large compared to the actual value.b.) The ice was dried so that the liquid on the ice wouldn’t be taken into effect in the calculation of the mass of ice which could produce error in the experiment because it would add more mass that wasn’t making an impact and would make the latent heat of fusion calculation too small.