Calculating Heat Absorbed & Estimating Power Rating of Electric Heater

In summary, the question asks how much power the electric heater is rated for and the answer is 60 watts.
  • #1
KevinO
3
0
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.
 
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  • #2
Hello KevinO. Welcome to Physics Forums.

KevinO said:
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
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
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
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.
 

1. How do you calculate the heat absorbed by an electric heater?

The heat absorbed by an electric heater can be calculated using the formula Q = mcΔT, where Q is the heat absorbed, m is the mass of the material, c is the specific heat capacity of the material, and ΔT is the change in temperature.

2. What factors affect the heat absorbed by an electric heater?

The heat absorbed by an electric heater is affected by factors such as the material and mass of the heater, the specific heat capacity of the material, and the duration of heating.

3. How do you estimate the power rating of an electric heater?

The power rating of an electric heater can be estimated by dividing the heat absorbed (calculated using Q = mcΔT) by the duration of heating.

4. Can the power rating of an electric heater change over time?

Yes, the power rating of an electric heater can change over time due to factors such as wear and tear, changes in the material or mass of the heater, and changes in the heating environment.

5. How accurate are estimates of the power rating of an electric heater?

The accuracy of estimates for the power rating of an electric heater depends on the accuracy of the inputs used in the calculation, such as the mass and specific heat capacity of the material, as well as the duration and conditions of heating. It is important to use accurate and consistent measurements to ensure a more precise estimate.

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