Hair dryer thermal energy

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Homework Statement



A hair dryer consists of a coil that warms air and a fan that blows the warm air out. The coil generates thermal energy at a rate of 600 W. Take the density of the air to be 1.25 Kg/m^3 and its specific heat capacity to be 990J/Kgk. The dryer takes air from a room at 20C and delivers it at a temperature of 40 C. What volume of air flows per second? The warm air makes water in the hair evaporate. If the mass of water in the air is 180g, calculate how long it will take to dry the hair if the HEAT required to evaporate 1g of water at 40C is 2200J.

Homework Equations



Density = M/v
Q = Mc(Change T)
P = Q/t

The Attempt at a Solution



I think this is a simple problem, however the answer sheet seems to disagree with me. The volume per second I calculated by simply finding the m/t which is basically P = mcT/t and assuming m/t = M; so I concluded that P = McT and, since i have the change in T and the specific heat capacity as well as P, I managed to calculate M. With this, I used Volume = Mass/Density and concluded that Volume/t = Mass/(Density x Time), where that equals the M I calculated beforehand. Supposedly, I should get 0.03/1.25 (where 0.03 = M). However, the answer key solves it by multiplying 1.25 x 0.03 or something? As for the second part, I don't think I have much trouble with it barring one little part. Basically I use cross-multiplication to find that 2200 x 180 = Energy needed to evaporated 180g of water, then divided that over the power rating of the dryer, so I should get something like 11 minutes. The answer key, however, solves it by dividing over 750. 750 is the product of 1.25 and 600. Why we chose to multiply those two numbers is beyond me. Anyways, I basically have no idea what to do to get that result. I tried thinking from the logic that 600 W releases 1.25 kg/m3 of air, but then I remembered that this 600W is the rate of ENERGY. So it's already Q/t. I'd appreciate any help, thanks!
 
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Answers and Replies

  • #2
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You're missing a volume flow rate for the fan, or at least omitted it for us.
 
  • #3
You're missing a volume flow rate for the fan, or at least omitted it for us.

That's exactly the givens for the question. You're meant to calculate the volume per unit time of the dryer. The answer sheet just says Density x (Change in Volume / Change in Time) and multiplies density with the mass per unit time for some reason? Is this an error or am I missing something?

Math-wise it makes little sense. Density = m/v, so v = m/Density. Therefore, (m/t) / density = V/t. Why do we have to multiply volume/t x density or even mass/t x density? They still won't give you the v/t.
 
  • #4
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Okay, my bad. "Dividing by 750?" Does sound like a "re-written" problem that failed to rewrite the answer key.
 
  • #5
Okay, my bad. "Dividing by 750?" Does sound like a "re-written" problem that failed to rewrite the answer key.

The answer key is directly from the book, so supposedly it has some merit. It's quite confusing to say the least. Is there any way to justify this answer? It seems like a very simple question, but it became needlessly complicated.
 
  • #6
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If it's an "nth" edition, n > 1, someone can decide to swap 600 for 750, and the graduate student slave labor fail to carry the value completely through the new answer key. There are no corrections for raising water temperature in the hair from body temperature of 37 C to 40, or allowing for heat input from the scalp, so the simplest interpretation is that the heater size in the dryer was changed from a previous edition.
 
  • #7
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It seems to me the book did the volume flow rate wrong, and you did it right. They should have divided by the density, not multiplied.

Chet
 

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