alt.energy.homepower

Thanks for the reminder; I had forgotten that and now everyone knows
that my days of teaching are many years in the past! LOL!!!

"David Williams" < @ > wrote in message
news: @ ...
>-> Every pound of ice that is 80 degrees cooler is 80 more btu's of
> -> cooling. -50* ice provides 82 btu's and liquifies, +30* ice provides
> =2=
> -> btu's and liquifies.
> -> Not all ice is created equal.
>
> Neglecting the difference bwween the specific heats of ice and liquid
> water, what you say is true *except for the words "and liquefies"*.
> Melting ice, from solid to liquid, absorbs *latent heat* from the
> surroundings, which, if I remember right, is about 50 calories per
> gram, or 100 BTU per pound. So, if you start with ice at -40 degrees (C
> or F - they're the same), you will have to provide 40 cal/g or 72
> BTU/lb just to warm it up to its melting point, then *another* 50 cal/g
> or 100 BTU/lb to melt it. So the ice will absorb a total of 90 cal/g,
> or 172 BTU/lb before it turns into liquid water at the freezing point.
> But if you start with ice that's already on the brink of melting, it
> will absorb only the latent heat, . 50 cal/g or 100 BTU/lb. So the
> ice that was initially very cold will absorb more heat, but by a factor
> of less than 2.
>
> dow