Heating the inside of a box thanks to a lamp

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Homework Help Overview

The discussion revolves around the heating of air inside a box using a lamp, specifically focusing on the time required to reach a certain temperature. The box's dimensions and the lamp's power are provided, along with relevant physical properties of air.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants explore the calculations related to the heating process, questioning the assumptions about energy loss and the characteristics of the lamp used. There are discussions about the impact of the box's design, such as the presence of holes and whether the air is stirred.

Discussion Status

Participants are actively engaging with the problem, raising questions about the assumptions made regarding energy loss and the heat capacities involved. Some guidance has been offered regarding the nature of the lamp and its heat emission characteristics, but no consensus has been reached on the best approach to calculate energy loss or the time required for heating.

Contextual Notes

There are constraints related to the physical setup of the experiment, including the presence of holes in the box and the lack of air stirring, which may affect heat transfer. The original poster's measurements and observations during the experiment also suggest discrepancies in expected heating times.

Rajo Yajo
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Homework Statement


The box's measurements are 277*120*195mm.
The lamp's power is 15W.
Trying to figure out how long it will take to heat the box so that it reaches a given temperature.
According to Wikipedia ρair=1,2kg/m3 and cp air=1004J*K-1*kg-1

Homework Equations


ΔU=m*cp*ΔTemperature
E=p*Δt

The Attempt at a Solution


If we suppose that the inside of the box is only air, we will have
vair=0.277*0.120*0.195=0.0064818m3
So, mair=vairair
mair=0.00777816kg
And,
ΔU=mair*cp air*ΔTemperature
We'll have,
ΔU≈7.8ΔTemperature
And if we assume there is no energy loss,
ΔU=E
So
7.8ΔTemperature=15Δt
Thus
Δt=0.52ΔTemperature.

The time I get seems to be too short.
Thanks in advance :wink:
 
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I think your answer is correct. The volume is very small - a cube around 20 cm in each dimension.
 
Is the box airtight?
 
Thank you for replying :)

Actually, there are five little holes on the box. They allow to put temperature sensor inside the box.

Is it possible to calculate the loss of energy due to these holes?

ps: When we do the experience, it takes around 30 seconds to gain 3°C.
 
If it's an incandescent lamp then most of the power is emitted as heat in the lamp itself. So you might need to allow for the heat capacity of the lamp and the fitting or wires?.

If the lamp is an LED then most (more?) of the power is emitted as light and that will be absorbed by the walls of the box. So you may need to allow for the heat capacity of the box walls. In addition some of the heat will be conducted and radiated to the outside.

Is the air in the box stirred so that the temperature recorded is representative?
 
Rajo Yajo said:
Is it possible to calculate the loss of energy due to these holes?

With difficulty. You would need to know the air flow rate into and out of the box. Easier to just seal them up.
 
Rajo Yajo said:
ps: When we do the experience, it takes around 30 seconds to gain 3°C.
Are you measuring the temperature of the air inside the box or of the box itself? What about the heat capacity of the box? Your original post asks about the box reaching "a given temperature."
 
It is an incadescent lamp, but I don't know whether I should consider its heat capacity to be E=P*Δt or use the heat capacity of tungsten.

The air inside the box is not stirred.
So I guess that the majority of the heat transfert is done by convection.

We are measuring the temperature of the air inside of the box (sorry for the ambiguity).
 
Pretty sure convection won't even start until the glass of the bulb is hot and convection can be quite a slow process. I would use the specific heat capacity of the glass part of the bulb. The filament gets white hot pretty quick so it's heat capacity will be very low.
 

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