Thermodynamics - Melting of Ice

AI Thread Summary
The discussion revolves around estimating the melting temperature of ice at the bottom of a 100m deep glacier, considering the pressure exerted by the ice above. The density of ice is noted as 0.92×10^3 kg/m^3, and its latent heat of fusion is 3.3×10^5 J/kg. A participant calculates the pressure at the glacier's base to be approximately 9×10^5 Pa, equating to about 100 atmospheres, which can lower the melting point of ice by roughly 1°C. There is confusion regarding numerical results and units, with a suggestion that a typo may have occurred in the density value. The conversation emphasizes the importance of correctly applying thermodynamic principles to solve the problem.
TheBaker
Messages
18
Reaction score
0

Homework Statement



The density of ice is 0·92×103 kg m−3 and its latent heat of fusion is 3·3×105 J kg−1.

Estimate the melting temperature of the ice at the bottom of a glacier which is 100m deep.

Homework Equations



L = Tc(S2 - S1)

The Attempt at a Solution



(S2 - S1) = 1208.8

I then tried to get somewhere with this, but kept going around in circles and couldn't work out how to incorporate the fact it's at the bottom of the glacier.

I'm guessing it's something to do with the pressure of the ice above, and worked out this was

p = F/A = 9×105

but, couldn't figure out what to do with it from there. Any help would be appreciated.
 
Last edited:
Physics news on Phys.org
Numerical results, unless they are pure numbers should be presented with units.
Are we speaking of 1208.8 sausages and 9×10e5 lumps of sugar?

A little bit of internet searching will tell you that a pressure of 100 atmospheres will lower the melting point of ice (generally 0°C) by about 1°C, that is to say not a lot.
Here's the reference
http://encarta.msn.com/encyclopedia_761570638/ice.html

So how much pressure is there at the bottom of a 100m thick glacier? Pressure in a medium like this is calculated as height*density (which gives us force per unit area). What does this mean approximately in terms of atmospheres (don't neglect the atmosphere itself).
 
Last edited by a moderator:
TheBaker said:
The density of ice is 0·92×10-3 kg m−3

This may be a simple typo. I suspect either m should be cm here, or the 10-3 should be 10+3
 
Sorry, that should indeed be +3
 
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
Back
Top