Air passes over an electrical heater at a steady rate of 2500cm^3 per second.

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Homework Help Overview

The problem involves calculating the heat absorbed by air passing over an electrical heater and estimating the power rating of the heater. The scenario includes specific parameters such as the volume flow rate of air, inlet and outlet temperatures, density of air, and specific heat capacity.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • Participants discuss the application of the equation Q=mcΔθ and question the calculation of mass based on the density of air. There are attempts to clarify the correct density unit and its impact on the calculations.

Discussion Status

The discussion has highlighted a potential miscalculation regarding the density of air, with participants providing guidance on the correct unit. Some participants have indicated that they resolved their confusion after reviewing the problem.

Contextual Notes

There is a noted discrepancy in the density of air, with participants pointing out that the original poster may have used an incorrect unit (kg/cm³ instead of kg/m³). This has implications for the calculations being discussed.

Student-
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Homework Statement



Air passes over an electrical heater at a steady rate of 2500cm^3 per second. The inlet temperature of the air is 20^oC ( 20 degrees celsius ) and the steady outlet temperature is 40^0C ( 40 degrees celsius)
a) What heat is absorbed by air passing over the heater in 2 hours
b) Obtain an estimate for the power rating of the heater
Density of air ( 1.2 kgcm-3)
Specific heat capacity of air : 1000 J-1Kg-1K-1


Homework Equations


So I think that I should use the equation Q=mcΔθ


The Attempt at a Solution


Here's what I did :
2500cm^3=2500g=2.5kg
Since 2.5kg is heated every second then 2hrs of heating would heat 2.5x7200 = 18,000kg of air.
ΔθOutlet-Inlet temperature = 40-20 = 20 degree celsius change in temperature,
Specific Heat capacity of air : 1000 Jkg-1K-1
Q=mxcxΔθ
= 18000x1000x20
=3.6x10^8
Which is wrong because the answer at the back of the book had 4.32x10^5. I've tried, but I really need some help :) Thanks



 
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Welcome to PF!

Hi Student-! Welcome to PF! :smile:

(try using the X2 button just above the Reply box :wink:)

Quickly dividing 4.32 by 3.6, my eye is drawn to …
Student- said:
Density of air ( 1.2 kgcm-3)

:wink:
 
I think you have mis- calculated the mass of air.
The density of air is 1.2kg/m^3 not 1.2kg/cm^3
 
technician said:
I think you have mis- calculated the mass of air.
The density of air is 1.2kg/m^3 not 1.2kg/cm^3

Thanks a lot :D I got out the question and I saw my error- :)
 


tiny-tim said:
Hi Student-! Welcome to PF! :smile:

(try using the X2 button just above the Reply box :wink:)

Quickly dividing 4.32 by 3.6, my eye is drawn to …


:wink:


Thanks I got it out eventually :D :)
 

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