What Is the Correct Internal Temperature of a Sealed Box with a Lightbulb?

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Homework Help Overview

The problem involves a sealed cubical box with a lightbulb inside, where participants are trying to determine the internal temperature when the light is on, given specific dimensions and material properties. The subject area includes heat transfer and thermodynamics.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to apply the heat transfer equation but questions the validity of their calculated temperature. They express confusion over the correct cross-sectional area to use in the calculations.
  • Some participants question the use of the cross-sectional area formula, suggesting that the total area should account for all sides of the cube rather than just one side.
  • Others suggest that the initial calculation may lead to an incorrect temperature due to the misunderstanding of the area involved in heat transfer.

Discussion Status

The discussion is ongoing, with participants exploring different interpretations of the problem. There is a suggestion that the original poster's approach may need adjustment based on the correct understanding of the area through which heat is transferred. No consensus has been reached yet.

Contextual Notes

Participants are working under the constraints of the problem statement and the properties of the materials involved, while also addressing potential errors in the setup of their calculations.

UMDstudent
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Homework Statement



A cubical box 22cm on a side is constructed from 1.3cm -thick concrete panels. A 100 W lightbulb is sealed inside the box. What is the air temperature inside the box when the light is on if the surrounding air temperature is 20 C?

Homework Equations



rate of heat transfer : heat / delta small t = (k*A*delta big T)/ L

Where : k = characterizes the material : (in this case its concrete) : 0.8 W/m K
L = Thickness of the panels : .013m
A = cross-section area = s^2 : (.22m)^2 = .0484

The Attempt at a Solution



100 W = ((0.8)(.22)^2(Tin - 20 C)) / .013m // the 20 C is the Tout

Tin = 53.5754 = 54 Degrees Celsius.

According to Mastering Physics this is the incorrect answer. To me it makes sense, and after reviewing my work, I cant' seem to find the error. Any help will be greatly appreciated.

-UMDstudent
 
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A cube has 6 sides.
 
Right but its asking for the cross sectional area of a cube. If i use A = s^6 : (.22)^6 = .000113 = very low number. Once this is plugged into the above equation, I get a HUGE number for degrees Celsius which makes no sense.

That HUGE number = 539,019 degrees Celsius.

-UMDstudent
 
Why use A=s^6 ?
The total area through which the heat passes is equal to the area of one side of the cube multiplied by 6.
Total area A =6 x 0.22² [and not 1 x 0.22² as you originally calculated]
The rest of your reasoning is fine.
100 = 0.8 x A(Δθ)/0.013

Which will give you a temperature difference 6 times less than the one you originally calculated, with Tin being correspondingly much lower.
 
Last edited:

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