Calculate the latent heat of fusion of ice

[Kara386]

Homework Statement

I'm given that for water near 0 degrees, dp/dT = $-1.4\times 10^7PaK^-1$. And that water has a specific volume of $1\times 10^{-3}m^3kg^{-1}$, while for ice the specific volume is $1.09\times 10^{-3}m^3kg^{-1}$. Calculate the latent heat of fusion L for ice, and estimate the mass of a skater who melts ice by skating on skates 20cm long and 0.1mm wide. Assume that the rink is at -3 degrees.

Homework Equations

The equation for heat of fusion I know is $Q=mL$, which I don't think is the way to go here. I did find out that $\frac{dp}{dT} = \frac{L}{T\Delta V}$, and I think that's the more useful equation.

The Attempt at a Solution

$\frac{dp}{dT}T\Delta V = L$

So I need to calculate $\Delta V$, but I'm not 100% sure how. If it's related to the 'specific volume', I've never heard of that quantity before and google had nothing useful to say about it...Unless $\Delta V$ is just the difference in specific volume between water and ice?
Also, since I'm multiplying by T and T=0, that whole equation would equal zero anyway, and it definitely shouldn't!
Thanks for any help :)

 

[drvrm]

Also, since I'm multiplying by T and T=0, that whole equation would equal zero anyway, and it definitely shouldn't!

you can use T in degree K to avoid zero/

 

[drvrm]

So I need to calculate ΔVΔV\Delta V, but I'm not 100% sure how.

i think you should take the difference in specific volume and calculate- the numbers will assure you about its meaning!

 

[SteamKing]

Homework Statement

I'm given that for water near 0 degrees, dp/dT = $-1.4\times 10^7PaK^-1$. And that water has a specific volume of $1\times 10^{-3}m^3kg^{-1}$, while for ice the specific volume is $1.09\times 10^{-3}m^3kg^{-1}$. Calculate the latent heat of fusion L for ice, and estimate the mass of a skater who melts ice by skating on skates 20cm long and 0.1mm wide. Assume that the rink is at -3 degrees.

Homework Equations

The equation for heat of fusion I know is $Q=mL$, which I don't think is the way to go here. I did find out that $\frac{dp}{dT} = \frac{L}{T\Delta V}$, and I think that's the more useful equation.

The Attempt at a Solution

$\frac{dp}{dT}T\Delta V = L$

So I need to calculate $\Delta V$, but I'm not 100% sure how. If it's related to the 'specific volume', I've never heard of that quantity before and google had nothing useful to say about it...Unless $\Delta V$ is just the difference in specific volume between water and ice?
Also, since I'm multiplying by T and T=0, that whole equation would equal zero anyway, and it definitely shouldn't!
Thanks for any help :)

The specific volume of a substance is that volume occupied by a unit mass of the substance. That's what the word 'specific' means in this context, whether it's referring to specific heat, specific volume, whatever.

The specific volume of liquid water is one of the key physical properties underlying the metric system. If that doesn't tell you what its value should be, then you can always look it up.

 

[Kara386]

you can use T in degree K to avoid zero/

Of course. Thank you!

The specific volume of a substance is that volume occupied by a unit mass of the substance. That's what the word 'specific' means in this context, whether it's referring to specific heat, specific volume, whatever.

The specific volume of liquid water is one of the key physical properties underlying the metric system. If that doesn't tell you what its value should be, then you can always look it up.

I should probably have realized that. In hindsight, the units provide a pretty good clue to the definition!

For the second part, I suspect that the skater is going to have to be incredibly heavy to melt ice through pressure. The pressure a skater of mass m exerts will be mg/A. And the heat it takes to melt the ice will be the heat required to raise the temperature to 0 plus the latent heat of fusion. So

$Q = mL + mcdT$

Where m is the mass of the ice to be melted rather than the mass of the skater... I suppose I could calculate for 1kg, or maybe work in heat per kg to melt ice?

Maybe that's the wrong approach entirely, and I should find the pressure required to lower the melting point of ice to -3 degrees. Using the Clausius-Clapeyron equation probably.

 

[Chestermiller]

At what pressure is the melting point of water -3 C?

 

[Kara386]

Oh, I got it! Integrating Clausius Clapeyron and rearranging a little bit. Thanks!