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

AI Thread Summary
The discussion centers on calculating the internal temperature of a sealed box containing a 100 W lightbulb, constructed from concrete panels. The initial calculation by a student yielded an internal temperature of approximately 54 degrees Celsius, which was deemed incorrect. Another participant pointed out that the cross-sectional area should account for all six sides of the cube, not just one, leading to a significant adjustment in the calculations. By using the correct total area, the temperature difference would be reduced, resulting in a much lower internal temperature. The correct approach emphasizes the importance of accurately determining the area for heat transfer calculations.
UMDstudent
Messages
33
Reaction score
0

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
 
Physics news on Phys.org
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:
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
TL;DR Summary: Find Electric field due to charges between 2 parallel infinite planes using Gauss law at any point Here's the diagram. We have a uniform p (rho) density of charges between 2 infinite planes in the cartesian coordinates system. I used a cube of thickness a that spans from z=-a/2 to z=a/2 as a Gaussian surface, each side of the cube has area A. I know that the field depends only on z since there is translational invariance in x and y directions because the planes are...

Similar threads

Replies
6
Views
2K
Replies
5
Views
2K
Replies
16
Views
2K
Replies
12
Views
2K
Replies
3
Views
4K
Replies
8
Views
1K
Replies
8
Views
3K
Back
Top