Brian Whatcott wrote:
On Sun, 02 Jul 2006 08:31:58 -0400, Jeff wrote:

Chris wrote:
For water, the heat of fusion is about 80 times specific heat...
So -70 C ice would have almost twice the cooling effect
as the same amount of barely frozen ice, right?
Wrong. You should look up the Heat Capacity of cold ice. Its only
0.5 BTU/lb-degree at freezing, but it goes down so that by -50 F its
only 0.4.

Since the Heat of Fusion is 144 BTU/lb, sub-cooling even 100 degrees
only adds a small amount of cooling capacity.

Hardly 'hadly effective'.
No, its hardly effective.

This is in error: using old CGS units
heat for fusion of ice is 80 cal/gm specific heat cap near 0degC is
1 cal/gm

Supercool to -40 deg C and its worth roughly another 40% of cooling
power cf. ice at freezing.

Brian Whatcott Altus OK

Sorry about the late reply - I've been out sailing.

You're making a common mistake. Although the specific heat of water
is 1 calorie per g-deg C, for ice its only about half that, or .5 cal
per g-deg C, or as I stated .5 BTU/lb-deg F. Thus, super cooling ice
add little cooling capacity.

On Sat, 08 Jul 2006 20:06:58 -0400, Jeff wrote:



This is in error: using old CGS units
heat for fusion of ice is 80 cal/gm specific heat cap near 0degC is
1 cal/gm

Supercool to -40 deg C and its worth roughly another 40% of cooling
power cf. ice at freezing.

Brian Whatcott Altus OK

Sorry about the late reply - I've been out sailing.

You're making a common mistake. Although the specific heat of water
is 1 calorie per g-deg C, for ice its only about half that, or .5 cal
per g-deg C, or as I stated .5 BTU/lb-deg F. Thus, super cooling ice
add little cooling capacity.


This time I checked a little more carefully among the welter of old
American Customary, CGS old scientific and SI units people have been
applying.
First, agree that ice has half the specific heat capacity of water
and agree that fusion of ice takes 80 cal/gm or 80 X 4.2 J/gm or
80 x 4.2 x 1000 J/kg
so -40 degC to melting point provides 20 cal or 20 X 4.2 J/gm or
20 X 4.2 X 1000 J/kg
and
ice to water provides 80 cal/gm or 80 X 4.2 J/gm or 80 X 4.2 X 1000
J/kg

The ratio in question is ( 80 + 20 ) / 80 = 125%
(NOT 140%)

The extra 25% of cooling effect may qualify as "little" or
"appreciable". You choose! :-)

Regards
Brian Whatcott Altus OK