The Angry Professor must walk 500 m from the coffee shop to reach her office. The coffee shop serves coffee at 140F, and the Angry Professor likes to drink her coffee when it reaches temperatures between 115F and 125F.
The outdoor temperature is 10F. Assuming the coffee is carried in an uninsulated, cylindrical container (height = 10 cm, radius = 4 cm), use Newton's Law of Cooling to determine how fast the Angry Professor must walk to reach her office in time to enjoy her coffee. Note that the heat transfer coefficient of coffee is approximately .6 W/Km2.
Thursday, January 31, 2008
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20 comments:
It was my understanding that there would be no math.
Um. Briskly?
The Angry Professor needs a travel mug.
The Angry Professor needs an assistant.
The Angry Professor needs a coffee pot.
Or at least a microwave.
Hmm... fast enough that we may need to take into account the induced wind chill effect. Now that's tricky...
If I had more time over lunch, I'd try the heat equation. Surely we can calculate at what rate you can sip from the surface of the coffee.
0.6 W per square kilometre? I think you'd better patent that mug as soon as you can!
This is a trick question!
This question doesn't factor in the potential of Angry Professor getting hijacked by a colleague, student, or other ne'er-do-well on the way to her office.
A metal travel mug.
The answer is Thermos.
If the temperature is 10, you'll walk fast enough.
The angry professor needs a coffeepot.
The Angry Professor needs a shot of Jack in that coffee to up the antifreeze properties.
The Angry Prof needs a heating mantle on a ring stand heating water in a round-bottomed flask.
What, doesn't everyone have one?
I can't answer the question because I missed that lecture because my sister was having surgery and she needed me to bring her her slippers. Could you hand-deliver me the notes please? And will this be on the exam?
Raises hand... What's the specific heat of coffee?
dT = 15F = 8.3K
To-Ta = 130F = 72K
Coffee area = 112pi cm^2,
Coffee mass = 160pi g,
Assuming a specific heat of 4 J/gK...
dQ = (4 J/gK)(160pi g)(8.3 K)
dQ = 17 kJ
dt = dQ/hA(To-Ta)
dt = 17 kJ/((0.6 W/Km^2)(112pi cm^2) (72K)(1m^2/10000 cm^2))
dt = 3 hours?
Was h supposed to be 0.6 W/K cm^2?
(which makes the answer 1 sec?).
Bah. I'm not a heat guy.
What kind of coffee shop serves take-out coffee in a cup 4 inches tall and less than 2 inches across?
Do you bring your own espresso cup?
What's the insulation factor of the mitten?
Speak English!!!
We don't use centimeters in Muskogee...
We don't measure heat loss that way...
We always use those good old inches...
In Muskogee we keep the metrics at bay.
Well I'm proud to ne an okie from Muskogee...
Shiiiiiiit...!!! That might make a good song!
I just saw something at the local gas station that has me intrigued. There's a sign for the Extra Energy Coffee in the Blue Pot. Here's an article on the web about that.
I wonder if AP's coffee habits would change with this coffee. She's certainly walk faster. Maybe to the point of phasing out of this reality.
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