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Measuring Heat

  • Thread starter KevinO
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  • #1
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the question states exactly:

Air passes over an electric heater at a steady rate of 2500 cubic centimeters per second. the steady inlet flow temperature of the air is 20 degree celcius and the steady outlet temperature is 40 degree celcius.
part a. what is the heat absorbed by the air passsing over the heater in 2 hours
part b. obtain an estimate power rating of the heater. is this too high or too low? explain.
[density of air = 1.2kg/cm^3; specific heat capacity of air= 1000Jkgcm^-3]




the equations i used were P=E/t; E=mcΔθ; ρ=m/V



for part a: After some substitutions i said P= (ρVcΔθ)/(t) where V/t = 2500cm^3s^-1
then i multiply my answer from that by 7200seconds and said that was the heat absorbed.

for part b: i just took the answer from P= (ρVcΔθ)/(t) and said that was the estimated power rating. my answers were P= 60MW and E=4.32 × 10^11 J

the text however says the answer was P= 60W and E=4.32×10^5 J. i assumed my answer was correct and the book had a mistake. when i showed my teacher however she said it was wrong to try again but i stuck.
 

Answers and Replies

  • #2
gneill
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Hello KevinO. Welcome to Physics Forums.

the question states exactly:

Air passes over an electric heater at a steady rate of 2500 cubic centimeters per second. the steady inlet flow temperature of the air is 20 degree celcius and the steady outlet temperature is 40 degree celcius.
part a. what is the heat absorbed by the air passsing over the heater in 2 hours
part b. obtain an estimate power rating of the heater. is this too high or too low? explain.
[density of air = 1.2kg/cm^3; specific heat capacity of air= 1000Jkgcm^-3]
The units attached to the specific heat capacity for air look mighty suspicious. Should be J/(K*kg) I'd think.
the equations i used were P=E/t; E=mcΔθ; ρ=m/V



for part a: After some substitutions i said P= (ρVcΔθ)/(t) where V/t = 2500cm^3s^-1
then i multiply my answer from that by 7200seconds and said that was the heat absorbed.
I think it would be helpful if you were to expand on that ad show more details of your work. Could be you've got a problem with unit conversions...
for part b: i just took the answer from P= (ρVcΔθ)/(t) and said that was the estimated power rating. my answers were P= 60MW and E=4.32 × 10^11 J

the text however says the answer was P= 60W and E=4.32×10^5 J. i assumed my answer was correct and the book had a mistake. when i showed my teacher however she said it was wrong to try again but i stuck.
Yup, the books answers look okay to me, too.
 
  • #3
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thank you for the welcome:
Yes the units for it was suppose to be J/(K*kg)
my solution was:
P=ρVc∆θ/t=[1.2kg/(cm^3)×2500cm^3×1000J/(kg*K)×20K]/1s=6.0×10^7 J/(s)
P*t= E = 6.0×10^7 J/s × 7200s = 4.32×10^11 J
 
Last edited:
  • #4
gneill
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20,792
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Okay, your density for air is not correct (I should have spotted that earlier). If air had a mass of 1.2 kg for every cubic centimeter (about the size of a sugar cube), we'd be swimming, not walking!

The density of air should be 1.2 kg per cubic meter :wink:
 
  • #5
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Thank you it works out great now.I checked the book to make sure i did not type the units wrong but the book has it with the wrong units. She should have point that out though instead of making me frustrated thinking i doing something wrong. thanks again.
 

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