Calculating Power of Melting Ice with Thermoelectric Heater/Cooler

[lastoftheligh]

Homework Statement

I'm trying to calculate the power required to melt a 57.65g mass of ice, using what my professor calls a "thermoelectric heater and cooler consisting of many thermocouples" (I think its really called a Thermopile, but I havn't heard him use the term.)

I took measurements regarding the voltage across the system, which is essentially the 'thermoelectric heater and cooler' pancaked in between 2 aluminum blocks on each side, each of which has had a hole drilled into it into which a temperature-sensitive resistor has been put in. I've used the constant current and the voltage measurements to obtain the temperature of these blocks over many periods of time, and also used the temperatures of the blocks to melt a piece of ice.

Now...


I'm supposed to use the latent heat of ice to determaine the amount of energy transferred.

Homework Equations

Now, I believe that all I need is to set the Latent Heat equation equal to the power equation...Q=mL=Pt

However, he specifically told me that I would need to use Q=MC(delta)T, and to (making sure I pay attention to the units) recognize that Q is in the units of Joules per second.

The Attempt at a Solution

Now, I havn't had any thermodynamics in a...very, very long time.

So I went and did some research and recalled that specific heat doesn't apply if one encounters a phase change, whereas latent heat is specifically for a phase change. So my initial assumption is that he's wrong, and I just need to relate latent heat to power.

But I know he's not wrong in this, so I must be making a mistake somewhere.

 

[Hootenanny]

Homework Statement

I'm trying to calculate the power required to melt a 57.65g mass of ice, using what my professor calls a "thermoelectric heater and cooler consisting of many thermocouples" (I think its really called a Thermopile, but I havn't heard him use the term.)

I took measurements regarding the voltage across the system, which is essentially the 'thermoelectric heater and cooler' pancaked in between 2 aluminum blocks on each side, each of which has had a hole drilled into it into which a temperature-sensitive resistor has been put in. I've used the constant current and the voltage measurements to obtain the temperature of these blocks over many periods of time, and also used the temperatures of the blocks to melt a piece of ice.

Now...


I'm supposed to use the latent heat of ice to determaine the amount of energy transferred.

Homework Equations

Now, I believe that all I need is to set the Latent Heat equation equal to the power equation...Q=mL=Pt

However, he specifically told me that I would need to use Q=MC(delta)T, and to (making sure I pay attention to the units) recognize that Q is in the units of Joules per second.

This should be plain old Joules, which you can convert to power later.

The Attempt at a Solution

Now, I havn't had any thermodynamics in a...very, very long time.

So I went and did some research and recalled that specific heat doesn't apply if one encounters a phase change, whereas latent heat is specifically for a phase change. So my initial assumption is that he's wrong, and I just need to relate latent heat to power.

But I know he's not wrong in this, so I must be making a mistake somewhere.

If the ice is below the fusion phase boundary (i.e. 273.15 at 1atm) then you are going to have to supply heat first to raise the temperature to 273.15 and then supply further heat to actually melt the ice. Therefore, you will need both latent and specific heat equations. However, take note of my comment above, Q has units of Joules (assuming that C takes standard values).

 

[lastoftheligh]

No, the ice is taken to be at the fusion phase boundary.

I am confused however. See, I had to listen to a half hour long talk (but this was over a month ago, and my notes don't make sense to me) - on why I had to use MC(delta)T.

But this /is/ taken to be at the fusion phase boundary, and is just involving normal basic physics at standard conditions

Secondly, on the 'plain old joules' thing...your right, it is in plain old joules. To convert to power...uh 1 W = 1 J/S ...so I would just divide it by the time it took to melt the ice?

Sorry if these questions are stupid or if they arn't showing enough work; most of what I'm doing is via spreadsheet, and this is the main conceptual part of it, I guess. But what I've been told to do and what I know of the concepts don't match up. If I'm just melting ice at the fusion phase boundary, and stop the moment all the ice is melted, I can't see why I'd use specific heat!

 

[Hootenanny]

What about the energy required to raise the temperature of the blocks of aluminum?

 

[lastoftheligh]

Oh, I didn't think of that. Now that you mention it, I think the original context WAS in terms of aluminum.

Cool! Difficulty one SOLVED! :)

You rock!


Anyways...the power required to heat the ice, I think - should be easy to calculate independent of the actual experiment, assuming ideal conditions and a specific mass m of ice, correct?

So Q=ML=(watts/s)

Watts=Joules/s =I^2 * R, both of which are constants in this case (constant current and resistance.)

However, actually that's the wrong track for ice; but combiined with specific heat I can use that for the blocks.

Hmm, for the ice...I can't use an electrical or mechanical definition. I think I'll have to stick with my original ideas, P=E/T

So...

Ice:

ML/<Deltat> = Power? WOOT

And for the blocks:

Q=Mc(of aluminum) * (delta)t =(I^2R)/(delta)t

Actually, I think I'm wrong there. The idea of dividing by delta-t is wonky. I'd have Q=Mc(of aluminum) = I^2R...

 

[Hootenanny]

Oh, I didn't think of that. Now that you mention it, I think the original context WAS in terms of aluminum.

Cool! Difficulty one SOLVED! :)

You rock!

Cheers!

Anyways...the power required to heat the ice, I think - should be easy to calculate independent of the actual experiment, assuming ideal conditions and a specific mass m of ice, correct?

So Q=ML=(watts/s)

This should be Watts*Seconds

Watts=Joules/s =I^2 * R, both of which are constants in this case (constant current and resistance.)

Are you sure that the resistance is going to be constant? What about when the aluminum blocks are 'warming up'? However, you could get around this by bringing the blocks up to temperature first, and then introducing the ice.

 

[lastoftheligh]

Actually, that could be true...it makes sense, but /boy/ would it mess everything up.

For example, for the 'temperature' column of my 1st spreadsheet (I took a zero-reading without any ice for about 5 minutes) - for the first 1000 ohm resistor (which was embedded in the aluminum) I used the equation:

(((B8 - .00547A) - 1000 ohms)/(1000 ohms * .00392))=T


where B8 is just a spreadsheet variable representing the voltage from the resistor, the Amps is the current, the ohms are the resistance, and the '.00392' is just a constant related to the resistant to temperature equation of the device we're using. So I have...

(((V-I)-R)/(.00392I))=T

Where T and V are variables, and I was presuming I and R to be constants.

So I have a column of V's, and I turned them into one of T's.

 

[Hootenanny]

I'm not quite sure what your calculating here. Perhaps this problem would be better addressed by an engineer...

 

[lastoftheligh]

Pardon, I gave the wrong equation. Its: T=(((V/I)-R)/(.00392R)).

It doesn't matter however, I know I'm doing the right stuff on this part, as best as I can with the equipment and setup at hand.

What I'd like to know is...I'm doing my writeup. I'm calculating the power used. Would it be the power to heat up both aluminum blocks + the power to melt the ice? So I add up the Q's, yes?

 

[Hootenanny]

What I'd like to know is...I'm doing my writeup. I'm calculating the power used. Would it be the power to heat up both aluminum blocks + the power to melt the ice? So I add up the Q's, yes?

Sounds about right to me, but I'd calculate the power required to increase the temperature of the blocks and the power required to melt the ice separately and then sum the powers.