Thermometer Inside Outside
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Thermometer Inside Outside
Cooking Filet Mignon/Tenderloin?
Last night I decided to go all out and buy a $50, 1.74 lb prime "butterflied" tenderloin/filet mignon for tonight's valentines day dinner.
I typically don't doubt my steak cooking talent, but then again I typically don't cook steak of this cost/quality/thickness. And I only have one..so there is no room for a screw up (as opposed to your cheaper $5.99-9.99/lb rib-eye from Safeway).
I should have all the neccessary spices/ingredients/equipment to work with. I have a good gas grill w/ thermometer,an oven w/broiler & griddle if neccessary. I like steaks the way they were meant to be eaten light seasonings, browned outside (not burnt), and medium rare inside. Any recommendations on cooking? I'm sure you'll all come up with great ideas, so it'll be hard to choose the best answer. Thanks a bunch.
your grilling these right?
from the food channel I learned that grilling the correct way is
(Based in a 2" thick butterflyed fillet)
Hi Heat! Heat up the grill real weel-with top down/on
Put filet on grill let it cook 2 minutes-then DON"T flip it. turn it 90 degrees to get the impressive grill mark-cook for another 2 minutes THEN FLIP IT-pull it off and let it rest 3-5 minutes before cutting/serving
do the same on the second side-cook it for two minutes then turn it 90 degrees.
I like my steak rare to medium-this basic rule of thumb cooks it right every time!
How to Care for a Leopard Gecko : Gecko Cage Thermometer
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Newton's Law of Cooling problem in Differential equations?
At 2:00 pm, a thermometer reading 80 F is taken outside, where the air temp is 20 F. At 2:03 pm, the temp reading yielded by the thermometer is 42 F. Later, the thermometer is brought inside, where the air is at 80 F. At 2:10 pm, the reading is 71 F. When was the thermometer brought inside?
the temperature and the time are both missing. how can it be solved?
The trick is to recognize that you know the total elapsed time.
Newton's law of cooling states that the rate of temperature change of a body is proportional to the difference between the ambient temperature and the temperature of the body:
dT/dt = - k*(T - Ta)
This is a separable equation, and the solution can be found by directly integrating to obtain:
T(t) = Ta + (To - Ta)*exp(-k*t)
where To is the temperature of the object at time t = 0
In this case, we can easily find the value of k by using the data given in the first part of the question:
42F = 20F + (80 - 20)F*exp(-k*3 min)
22/60 = exp(-k* 3min)
(1/3min)*ln(22/60) = -k
k = 0.3344/min
Now, let's assume the thermometer was left outside for some unknown time t, and then was brought back into the room and at 2:10pm we know the temperature of the thermometer was 71F. Let the time elapsed between when the thermometer was brought back inside and 2:10pm be delta-t, then we know that t + delta-t = 10 min (i.e., the total elapsed time was 10 minutes).
When the thermometer is brought back inside, a new thermal adjustment occurs according to Newton's law of cooling, but now the initial temperature depends on how long it was left outside. The new equation governing the temperature as a function of time is:
T(t + delta-t) = 80F + ((20 + 60exp(-kt)) - 80)F * exp(-k*delta-t)
T(t + delta-t) = 80F + (60exp(-kt)) - 60)F * exp(-k*delta-t)
We know that at t+delta-t = 10min, T = 71F, so:
71F = 80F + (60exp(-kt)) - 60)F * exp(-k*delta-t)
9/60 = (1-exp(-kt))*exp(-k*delta-t)
But t + delta-t = 10 min, so delta-t = 10 min - t. Plugging this into the above equation, we have:
3/20 = (1-exp(-kt))*exp(-k*(10 min - t))
Remember that exp(a + b) = exp(a) * exp(b), so:
3/20 = (1-exp(-kt)) * exp(-10min*k) * exp(k*t)
3/20 = (exp(k*t) - 1) * exp(-10min*k)
(3/20)*exp(10min*k) = exp(k*t) - 1
1 + (3/20)*exp(10min*k) = exp(k*t)
t = (1/k)*ln[1 + (3/20)*exp(10min*k)]
Plugging in the value for k we obtained above, we find that:
t = 4.96 min, so the thermometer was brought back inside at 2:04.96 ~= 2.05pm
