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Topic: specific heat and formula (10 msgs / 284 lines)
1) From:
Afternoon.  Does anyone know the specific heat of green coffee?  Could only find brewed coffee (same as water) on the web.  
Does anyone understand the effect of mass on the specific heat formula? For example, if it takes 1000 joules to heat a 1/4 of coffee to 1st crack in 13 minutes, does it take 2000 joules to heat a 1/2 lb to 1st crack in 13 minutes?
Any thoughts on how to figure out the relationship of joules or calories in the formula to watts in an electric heating element?

2) From: Randall Nortman
On Thu, Jun 21, 2007 at 10:35:25AM -0700, thirddayhomeroaster wrote:
This job calls for a physicist!  Step back, everyone, help is here!
DISCLAIMER: I was a physics major at U of Chicago back in the day.
These days, I play with computers.  Not a real physicist, but I can
handle something simple like this.
Joules measure total energy.  Watts measure power.  Power equals
energy divided by time.  Conveniently, one watt equals one
joule per second.  So to transfer 1000 joules of heat energy into the
coffee with a heater power of one watt would take 1000 seconds.  If
the heater power is 1000 watts it would only take one second.  If
heater power is 500 watts, it would take two seconds, and so on.
P = E/t  (power = energy / time)
t = E/P  (time = energy / power)
E = Pt   (energy = power x time)
Calories are also a measure of energy.  One calorie = 4.2 joules.
However, when you read the nutrition label on the side of your
breakfast cereal, what they call a "Calorie" is actually a kilocalorie
(kcal), or 1000 calories.  Confusing, but there you have it.  Anybody
who knows what they're talking about will always spell food-related
calories as "Calories" with a capital "C", or else be explicit and say
"kcal" or "kilocalories" if it's an academic paper.
Now that simple answer has limited relevance to what you're trying to
figure out, for two main reasons.  First, the relationship between
heat transfer to the bean and resulting bean temperature is non-linear
at several points in the process, due to endothermic (heat-consuming)
and exothermic (heat-producing) chemical reactions.  There are
substantial non-linearities especially around first and second crack.
Secondly, only a small fraction of the heater power actually is
transferred to the beans, particularly in a hot air roaster, where the
majority of the heater power exits the roaster as hot exhaust air.  A
roaster that recirculates air will lose a lot less heat.  I suppose
that drum roasters that don't pull/push air through the beans at all
also lose less, though they still lose heat through the roaster walls,
plus the roaster itself sucks up heat as it comes to temperature.
So, if you want to do this right, step one is to determine the
transfer efficiency of the roaster; i.e., what fraction of the heater
power is transfered to the beans.  This will likely depend on air
flow, ambient temperature, roaster temperature, and bean temperature,
and so it will change through the roast.  Second is to figure out what
all the endo- and exothermic reactions are that occur during roasting,
along with the total reaction mass for those reactions and the
reaction rates at various temperatures.  Then you're ready to put that
all into one enormous equation and get to work integrating it over
time.  I suggest that computer-assisted numerical integration methods
would be the best approach in a situation like this, as there's not
enough paper or chalkboard space in a typical university to hold all
your scratchwork if you want to do it symbolically.
Good luck with that.

3) From: Greg Scace
Nah - any ol mechanical engineer who works with heat transfer will do.
Specific heat is the amount of heat required to change the 
temperature of a unit mass of substance  by1 degree.  Since we like 
to think of watts (power), it is convenient to thing of specific heat 
in terms of SI units.  Just fyi,  In SI units, specific heat is given 
in joules per kilogram degree C.  As mentioned before, power is 
energy per unit time - watts is joules per second.  Specific heat has 
no time component in it, so we're just talking about energy required 
to raise the temperature of the coffee by a certain amount.
Your example is correct.  It takes twice the amount of energy to heat 
twice the amount of substance to the same temperature.   Since you 
are interested in doing so in the same time interval it will take 
twice the power TRANSMITTED TO THE COFFEE.    The answer to your 
question is simple, but the real world question that i think you 
might be asking  (Do I need twice as big of a heating element to 
roast twice as big a load of coffee in the same time?) has a more 
complicated answer that needs to take into account changes in the 
efficiency of heat transfer from the heating element to the 
coffee.  Some areas that muddy the waters are the amount of coffee 
that is in physical contact with the drum (if drum roasting), changes 
in heat transfer to a larger or smaller mass from air convection in 
the drum, and others.
At 02:06 PM 6/21/2007, you wrote:

4) From: Greg Scace
Oh, I forgot to mention that you might be able to actually find the 
specific heat of coffee if you search for it on the net (I didn't 
try).  A first approximation could be to use the specific heat of 
cellulose, if you can't find a real value.  Then you need to know the 
amount of temperature rise from the initial temperature of the coffee 
(prolly room) to first crack, the mass of the coffee (use kg and 
Degrees C so you don't screw it up), and the time.  Power required is 
the specific heat multiplied by the temperature difference, divided 
by the time in seconds.   Then look at the heating element 
specification.  The efficiency of the system is the ratio of the 
power consumed by the coffee divided by the power produced by the 
heating element.  I think that is the relationship you asked for in 
your second paragraph.  I should point out that that relationship 
holds only for that amount of coffee in the roaster for the reasons I 
mentioned in my other email.
At 02:06 PM 6/21/2007, you wrote:

5) From: Dan Bollinger
Technically, the mass effect is linear. If it takes 1000 joules (about 1 BTU) to 
heat 1/4 pound of coffee to 1st crack that is what it takes regardless of if it 
is 13 seconds, 13 minutes or 13 years.  If you double the mass you double the 
energy required, all things being equal.
Practically, the above is untrue. You still have to deal with heating the 
roaster (which may very well require more heat than heating the beans, but is a 
lesser proportion than heating twice the amount of beans). And, you still have 
to deal with losses to the environment over time.

6) From: Stan Klonowski
I am relatively new here, but have a lot of experience involving water
and heating.  So...
I am going to go out on a limb and say that water content is going to be
the largest influence on the heat capacity of coffee.  Assuming
cellulose and coffee are fairly similar the heat capacity can be up to
1.3 times that of pure water.  Taking into account that water removal is
a key to roasting and the specific heat of the phase transition between
water and steam is higher than just raising the temp. one degree it
starts to get pretty sticky.  
Here is a reference if you want to read the ugly details:http://www.nature.com/nature/journal/v176/n4471/pdf/176083a0.pdfJust for the record I love the idea of trying to calculate roast
parameters theoretically, but as a working scientist I can tell you
there is a lot of empirical observation that also needs to be done.  
Let me know off list if the PDF does not work and you would still like
to see it. 

7) From:
I'm new too-2 days now.  I appreciate the response.  The link takes me to a screen that asks me to log in/buy the article.  Are you able to select adobe as your printer and email it that way?  I'd love to see what it has to say. I'm trying to beef up my roaster to the most that can be roasted on 120v.
 ---- Stan Klonowski  wrote: 

8) From: Dan Bollinger
Stan, You'll be interested in the articles from Association Scientifique 
Internationale du Café. http://www.asic-cafe.org/Dan

9) From: David Liguori
Basically, yes.  Heat is an "extensive" parameter, meaning if you double 
the system the heat content, or heat required to accomplish a reaction 
or change in temperature (both of which are happening in raising beans 
to first crack, as well as vaporizing water) also doubles.  Specific 
heat is an "intensive" parameter, expressed in joules per kilogram, 
joules per mole, etc.
Practically speaking, though, the heat transfer mechanism is way too 
complicated to calculate, meaning it is probably hopeless to try to 
correlate the rate at which heat is produced by the element with the 
heat required to roast your beans.  Often in fluid bed roasters, adding 
more bean mass makes the temperature rise faster, due to slower air 
flow.  Therefore, the specific heat of the beans is not likely to be of 
any practical use, which is probably why you're not finding it anywhere. 
  If you were really curious it's fairly easy to measure: just heat a 
known amount of beans to a known temperature, add them to a known mass 
of water in an insulated container like a Styrofoam cup, then measure 
the temperature when it stops changing.  This does not take into account 
the heat required to vaporize the water in the beans.  Once you're into 
cracks, of course, you have an endothermic reaction, further 
complicating things.
thirddayhomeroaster wrote:

10) From: Floyd Lozano
The problem will be solved soon with robots blasting fast
moving beans arranged single file with laser beams, so I wouldn't worry
about it.
On 6/22/07, David Liguori  wrote:

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