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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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