Laten heat of fusion calculation

[trying_mybest]

Homework Statement

I'm supposed to calculate the latent heat of fusion of water from the given experimental data.

6 ice cubes, each weighing 4 g, were used to bring the temperature of a 0.25-L water from 23.3˚ C to 0˚ C. The temperature of each ice cube is assumed to be -15˚ C.

Relevant Equations

Not sure if these are the correct equations to use:

Q = m * Cv * ∆T
Q = m * Lf

Q1 = m(ice) * Cv * ∆T = amount of heat the ice absorbs
Q2 = m(H2O) * Lf = amount of heat the water loses

• Q1 = (4*8 g) * (2198 J/g*K) * (0-(-15)) = 1.055 * 10^6 J
• Q1=Q2
• Q2 = 1.055 * 10^6 J = m(H2O) * Lf
• Lf = 1.055 * 10^6 J / m(H2O) = 1.055 * 10^6 J / (0.25 L * 1000 g/1L) = 4220 J/g

My answer is really far off from the actual Lf of water (334 J/g).

 

[TSny]

Which part of the system undergoes a phase change, the ice or the 0.25 L of water?

 

[trying_mybest]

Which part of the system undergoes a phase change, the ice or the 0.25 L of water?

The ice does the phase change I believe.

 

[TSny]

The ice does the phase change I believe.

Yes. So, the heat associated with the phase change should be part of the heat gained by the ice; that is, part of Q1.

How would you express Q2, the heat lost by the 0.25 L of water?

(I don't get a decent result for L_f. But let's see what you get.)

 

[TSny]

Do the numbers given in this problem represent the actual data of an experiment that you performed? Did all 6 ice cubes melt completely? Was the final equilibrium temperature of the system actually 0 ^oC?

 

[trying_mybest]

Because I don't have a thermometer to measure it accurately, I'm making the assumption that the final temperature is 0˚ C since that's when the ice stopped melting.

Also I redid the experiment. The initial temp of the water was 25.6 ˚C. I ended up using 11 ice cubes before the next one was barely melting.

I guess my main issue is figuring out the setup to figure out Lf.

 

[TSny]

Were you asked to add just the right amount of ice so that all of the ice melts, but any additional ice would not melt? Or, was this your own idea? You are right that if you could do this, then the final temperature of the system would be 0 ^oC. But this could be tricky to carry out accurately.

What type of material is the container made of? Styrofoam would probably be best. Otherwise, there could be a significant amount of heat lost by the material of the container as the container cools along with the water. This would need to be taken into account.

Are you just assuming the initial temperature of the ice is -15^oC? This assumption might not be valid.

I don't know if you are supposed to follow a particular procedure that was given to you. If so, it might help to post the procedure. But, if you are free to decide for yourself how to proceed, you might consider doing the following. Put a bunch of ice cubes in a glass and let the glass sit in your room for a while until a significant amount of ice has melted. Since the ice in the glass is now in the process of melting, what must be the (approximate) temperature of the remaining ice cubes in the glass? Add some of these cubes to the water and measure the final equilibrium temperature. Hopefully, you have a way to measure the mass of the water before and after the ice is added so that you can deduce the mass of the ice that was added. It appears that you can measure the initial temperature of the water to the nearest tenth of a degree or so. Therefore, you should be able to measure the final equilibrium temperature to a similar accuracy. Then, set up the "heat gained = heat lost" equation to find the latent heat of fusion of the ice.

 

[Chestermiller]

The final temperature of the system is equilibrated at 23.3 C.

 

[trying_mybest]

This is the procedure.

"Take a large glass or a bowl and fill it with 0.5 liter of water at room temperature, (or let it sit until it is at room temperature)

Then estimate how many ice cubes (from your freezer) you think it will take to bring the glass or bowl with 1/2 liter of water down to 0 degrees Celsius. Assume the ice cubes are at -15 C. If you don't have a scale to measure the mass of 1 ice cube,, measure the ice cube and using the known density of ice determine the approximate mass. Then add enough ice cubes, until the ice added no longer melts. (This will happen when the water temperature is at 0 degrees C). From the experiment determine the latent heat of fusion of water."

The experiment was to be done at home with whatever materials we have. I used 0.25 L of water instead because I don't have a container that can hold that volume + the amount of ice. The container was plastic.

 

[TSny]

This is the procedure.

"Take a large glass or a bowl and fill it with 0.5 liter of water at room temperature, (or let it sit until it is at room temperature)

Then estimate how many ice cubes (from your freezer) you think it will take to bring the glass or bowl with 1/2 liter of water down to 0 degrees Celsius. Assume the ice cubes are at -15 C. If you don't have a scale to measure the mass of 1 ice cube,, measure the ice cube and using the known density of ice determine the approximate mass. Then add enough ice cubes, until the ice added no longer melts. (This will happen when the water temperature is at 0 degrees C). From the experiment determine the latent heat of fusion of water."

The experiment was to be done at home with whatever materials we have. I used 0.25 L of water instead because I don't have a container that can hold that volume + the amount of ice. The container was plastic.

OK, thanks for posting the procedure. That really helps. You can ignore the alternate procedure of my previous post. Don't expect to get an accurate result in this experiment.

So, you are having difficulty setting up the equation for finding L_f. You have the right idea that Q1= Q2, where Q1 is the total heat gained by the ice, and Q2 is the total heat lost by the water.

Note that Q1 must take into account the heat gained by the ice to increase its temperature from -15 ^oC to 0 ^oC as well as the heat gained by the ice in order to melt completely. So, the expression for Q1 will have two terms. Can you see what these two terms look like?

Q2 is the heat lost by the water as it cools. This is just heat associated with a temperature change. How would you write an expression for Q2?

 

[trying_mybest]

OK, thanks for posting the procedure. That really helps. You can ignore the alternate procedure of my previous post. Don't expect to get an accurate result in this experiment.

So, you are having difficulty setting up the equation for finding L_f. You have the right idea that Q1= Q2, where Q1 is the total heat gained by the ice, and Q2 is the total heat lost by the water.

Note that Q1 must take into account the heat gained by the ice to increase its temperature from -15 ^oC to 0 ^oC as well as the heat gained by the ice in order to melt completely. So, the expression for Q1 will have two terms. Can you see what these two terms look like?

Q2 is the heat lost by the water as it cools. This is just heat associated with a temperature change. How would you write an expression for Q2?

This is what I think it should be:

Q_{heat H2O lost} = Q_{ice to reach 0 C} + Q_{melt ice}

• Q_{heat H2O lost} = m_H2O * C_v,H2O * ∆T_H2O
  • Q_{heat H2O lost} = [0.25 L * 1000 g/1L] * (4.184 J/g*K) * (0-23.3) = -24,372 J
• Q_{ice to reach 0 C} = m_ice * C_v,ice * ∆T_ice
  • Q_{ice to reach 0 C} = (4 * 11) * (2.108 J/g*K) * (0-(-15)) = 1391 J
• Q_{melt ice} = Q_{heat H2O lost} - Q_{ice to reach 0 C}
  • Q_{melt ice} = -24,372 J -1391 J = -25,763 J
  • Q_{melt ice} = m_ice * L_f
    • L_f = Q_{melt ice} / m_ice = 25,763 / 44 = 586 J/g

 

[TSny]

Your calculation is almost correct. The formula $Q = mc \Delta T$ gives the heat gained by a substance. So, your answer of -24,372 J represents the heat gained by the water. The negative answer means that heat was lost. The amount lost is +24,372 J.

So, in your equation $Q_{\rm melt \, ice} = Q_{\rm heat \, H20 \, lost} - Q_{\rm ice \, to \, reach \, 0 \, C}$, the first term on the right should be +24,372 J. Otherwise, your method looks correct. Was the initial temperature of the water 23.3 ^oC or 25.6 ^oC?

 

[trying_mybest]

My first attempt, the initial temp was 23.3, but then the second attempt was 25.6. I will make that correction on my calculation.

Thank you so much for all the help!

 

[TSny]

My first attempt, the initial temp was 23.3, but then the second attempt was 25.6. I will make that correction on my calculation.

Sounds good.

Thank you so much for all the help!

You are welcome.