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Group # 5 2BPH Date Performed: February 22, 2013Members: Date Submitted: February 29, 2013(26) Lee, Sharmaine Margaret (29) Monsalud, Miguel

(27) Loterte, Edwin Jr. (leader) (30) Montalbo, Lora Jean

(28) Lumbao,Jicah Mae (31) Nano, Lizette

Experiment 8

Latent Heat of Fusion

Abstract:

The objective of the experiment is to calculate the heat fusion of ice and compute for thepercentage error using the standard of 80 cal/g. Heat of fusion is the change in enthalpy resulting

from heating a given quantity of a substance to change its state from a solid to a liquid. It is the

needed heat to change the phase of a substance without change in the temperature. Each group is

provided with a triple beam balance, calorimeter with a specific heat of 0.215 cal/g Co, mercury

thermometer , hot water and ice. The students

Question & Answer:

1.  Since fusion and melting do not result into a temperature change, where does the energy

go?

When a solid has reached its melting point, additional heating melts the solid without a

temperature change. The temperature will remain constant at the melting point until the entiresolid has melted. The amount of heat needed to melt the solid depends only on the mass of the

solid. We have: Q = mLf  

Where:

Q is the amount of heat absorbed by the solid, m is the mass of the solid and

 Lf is the latent heat of fusion measured in cal/g (to fuse means to melt)

Ice will be added to a calorimeter containing warm water. The heat energy lost by thewater and calorimeter does two things: melts the ice and warms the water formed by the

melting ice from zero to the final temperature.

Heat lost by warn water + heat needed to warm water which was once ice + heat neededto melt ice = 0

MwCw(Tf -Ti) + MiceCw(Tf + MiceLf -0) = 0 * Note: The mass of the melted water is the same as the mass of the ice.where

 M w = mass of warm water initially

in calorimeter

 M ice = mass of ice and water from melting

C w = specific heat of water

 Lf  = heat of fusion of ice

Ti = initial temperature water

Tf  = equilibrium temperature of mixture

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2.  What source of error is present in this experiment that was not present in the experiment

on specific heat capacity?

Stirring: It is important to stir the water and ice mixture to ensure that the temperature

throughout the water is uniform. Not stirring the ice and water mixture causes the final

temperature to be too warm and gives an experimental value of the Latent Heat of Fusion that istoo low.

Thermometer: The SS thermometer should not come into contact with the calorimeter. This

contact causes the final temperature to be too warm and gives an experimental value of theLatent Heat of Fusion that is too low.

Drying The Ice: If the ice is not dried there will be water at 0oC on the ice. The added water will

contribute to the final mass of liquid but it will not gain the amount of heat that an equivalentamount of ice would gain. The initial temperature of the water in the calorimeter will not have to

drop as far. Hence the final temperature will be too high. The result will be an experimental

value of the Latent Heat of Fusion that is too low

3. 

How much heat is absorbed by an electric refrigerator in changing 2.00 kg of water at15.0oC to ice at 0.0

oC?

4.  Determine the resulting temperature when 150.00g of ice at 0.0 oC is mixed with 300.00 g

of water at 50.0oC

Following: QH2O + Qice + mLf = 0

MwCw(Tf -Ti) + MiceCw(Tf -0) + MiceLf = 0 

(300.00g)(1cal/gCo)(X-50.0C

o) + (150.00g)(1cal/gC

o)(X-0.00C

o) + (150.00g)(80 cal/g) = 0

(300.00 cal/Co)X - 15000cal + (150.00 cal/C

o)X - 0 + 12000cal = 0

(450.00 cal/Co)X – 3000cal = 0

(450.00 cal/Co)X = 3000cal

X = 7 oC