latent heat of fusion , how do I put this in a formula to find heat of fusion?

Question

100mL water @ 21 degrees C, add an ice cube, then there is
125mL @ 7 degrees C after ice melts

Answer 1

The two formulas you will need to employ here are:
Q = mc(delta T)
and
Q = mL
where m is the mass, c is the specific heat, (delta T) is the change in temperature, L is the latent heat of fusion, and Q is the heat energy.

You know that the heat gained by the ice equals the heat lost by the water. You know that at equilibrium both the "ice" and the "water" are not at the same temperature, and in this case they are both now in the liquid state.

Since you know the final temperature of the mixture as well as the initial temperature of the water you can calculate the heat lost by the water and set it equal to the heat gained by the ice.
Find the heat lost by the water using
Q = mc(delta T),
set it equal to the heat gained by the ice,
Q = mL + mc(delta T)
notice that this has two parts...first the ice must melt, the the resulting water's temperature must be raised to 7 degrees.

I think that it is assumed that you know the specific heat of water (4.184 J/g degree C) as well as the density (1 g/mL) in this problem.

The actual value of the latent heat of fusion of water is about 334 Joules per gram, you can use this to check your answers.
(make sure you use the correct values for the masses of the ice and water, otherwise you will get the wrong answer)

Answer 2

Latent Heat Of Fusion Equation

Answer 3

Heat Of Fusion Formula

Answer 4

This Site Might Help You.

RE:
latent heat of fusion , how do I put this in a formula to find heat of fusion?
100mL water @ 21 degrees C, add an ice cube, then there is
125mL @ 7 degrees C after ice melts

Answer 5

Hi your basic question is wrong.instead of 125 ml it should be 115 ml then it make sense
Remember the latent heat of water is 1 that means 1 calorie of heat given or taken away from 1 gram of water will heat or cool the water by 1 degree celcious.
Now let us make the formula
total heat contained in water above 0 degree C is
100 gm x 21=2100 cal
now heat lost is heat gained ,if x is the latent heat of fusion of ice and we have 15 gm of ice then the for mula will be written like this
2100={[100+15].7}+x.15.or
2100=905+15.x or
2100-905=15.x or
1205=15x or
x=1205/15 =80.3appox
The latent heat of fusion of ice is 80 calories so we are all right here.I sincerely hope this helps you.

Answer 6

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