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Re: Engineers Diet

To: <DSRGR@aol.com>, <vintage-race@autox.team.net>, <9laser3@bright.net>,
Subject: Re: Engineers Diet
From: "Roger Sieling" <Rogsie@telesistech.com>
Date: Thu, 09 Aug 2001 16:55:32 -0400
But why would anyone polute hot wings w/ bleau cheese?

>>> "Paul" <9laser3@bright.net> 08/09/01 03:51PM >>>

  I thought the celery was just to shovel up more blue cheese dressing when
you run out of wings!

Paul
OROC
-----Original Message-----
From: Roger Sieling <Rogsie@telesistech.com>
To: DSRGR@aol.com <DSRGR@aol.com>; vintage-race@autox.team.net
<vintage-race@autox.team.net>; haynes386@netzero.net <haynes386@netzero.net>
Date: Thursday, August 09, 2001 12:12 PM
Subject: Re: Engineers Diet


>To go with their bloody Mary? But I could never understand why anyone would
>wnt to ruin the taste of that or Bufalo wings with celery.
>
>>>> Mark Haynes <haynes386@netzero.net> 08/08/01 11:27PM >>>
>Sorry Jeff, but as you may have forgotten, it's one calorie to raise one
gram
>of
>water one degree C, however, it's one Kilocalorie (Cal) in human consumptin
>terms. this means that it takes .37 Cal to raise 1 gram  of mostly water
>dessert
>by 37 C.....By the way, why would anyone eat celery for dessert??
>Mark Haynes
>
>DSRGR@aol.com wrote:
>
>> Engineers Diet
>> We all know that it takes 1 calorie to heat 1 gram of water 1 degree
>> Celsius.
>> Translated into meaningful terms, this means that if you eat a very cold
>> dessert (generally consisting of water in large part), the natural
>processes
>> which raise the consumed dessert to body temperature during the digestive
>> cycle literally sucks the calories out of the only available source, your
>> body fat. For example, a dessert served and eaten near 0 degrees C (32.2
>> deg.F) will in a short time be raised to the normal body temperature of
37
>> degrees C (98.6 deg. F).  For each gram of dessert eaten, that process
>takes
>> approximately 37 calories as stated above.  The  average dessert portion
is
>> 6 oz, or 168 grams. Therefore, by operation of thermodynamic law, 6,216
>> calories (1 cal./gm/deg. x 37 deg. x 168 gms) are extracted from body fat
>as
>> the dessert's temperature is normalized. Allowing for the 1,200 latent
>> calories in the dessert, the net calorie loss is approximately 5,000
>> calories.  Obviously, the more cold dessert you eat, the better off you
are
>>
>> This process works equally well when drinking very cold beer in frosted
>> glasses.  Each ounce of beer contains 16 latent calories, but extracts
>1,036
>> calories (6,216 cal. per 6 oz. portion) in the temperature normalizing
>> process.  Thus the net calorie loss per ounce of beer is 1,020 calories.
>>
>> It doesn't take a rocket scientist to calculate that 12,240 calories
(12oz.
>> x 1,020 cal./oz.) are extracted from the body in the process of drinking
a
>> can of beer. Frozen desserts, e.g., ice cream, are even more beneficial,
>> since it takes 83 cal./gm to melt them (i.e., raise them to 0 deg. C) and
>an
>> additional 37 cal./gm to further raise them to body temperature. The
>results
>> here are really remarkable, and it beats running hands down.
Unfortunately,
>> for those who eat pizza as an excuse to drink beer, pizza (loaded with
>> latent calories and served above body temperature) induces an opposite
>> effect. But, thankfully, as the astute reader should have already
reasoned,
>> the obvious solution is to drink a lot of beer with pizza, and follow up
>> immediately with large bowls of ice cream. We should all be thin very
soon
>> if we adhere religiously to this cold pizza, cold beer, and ice cream
diet.
>>
>> Jeff Vance
>> Ohio Department of Transportation
>
>
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