Conversion of work to heat: Getting the Sun's temperature inside a braking block

AI Thread Summary
The discussion revolves around estimating the mass of train brakes needed to prevent overheating and the number of photons emitted by a light bulb. For the brakes, the initial calculations incorrectly suggested an absurd temperature due to a misunderstanding of energy conversion. Clarifications indicated that the focus should be on the energy (K) entering the brake block and how to calculate the necessary mass to avoid exceeding the melting point. Additionally, the light bulb's power output and photon emission were discussed, with suggestions to verify the wattage and energy conversion efficiency of different bulb types. Overall, the conversation highlights the importance of accurate energy and temperature calculations in physics problems.
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Homework Statement



Estimate

(i) the mass of the brakes of a train needed to stop the brake block temperature reaching melting point when the brakes are applied

(ii) the number of photons emitted per second by a light bulb

An order of magnitude estimate is sufficient.

Homework Equations



The Attempt at a Solution



For i),

I assumed that all of the kinetic energy of the train is converted to thermal kinetic energy of the material.

The thermal energy is on the order of kT.

I assumed that the speed of the train is 30 m/s and that the mass of the train is 2000 kg.

So, kT = 0.5mv2
This gives T = 3.2*1030 K.

This is clearly absurd!


For ii),

power of a typical light bulb = I2R = (50 mA)2(1.5 ohms) = 380 mW.

And energy of one light photon = hc/lamba = 2.9*10-17 J.
(assuming lamba = 700 nm).

So, number of photons = power of light bulb / energy of one photon = 1.3*1012 s-1,

which is a sensible number.

I'd be grateful to you to anyone who suggests a much better way.
 
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hasan_researc said:
The thermal energy is on the order of kT.

You mean kinetic energy? "kT" is something I don't recognize.

I assumed that the speed of the train is 30 m/s and that the mass of the train is 2000 kg.

So, kT = 0.5mv2

OK, you do mean kinetic energy.
This gives T = 3.2*1030 K.

No, this gives K=3.230 joules. This will be the "heat" (Q) that enters the block. Energy is not temperature.

Now you have to find out what mass is necessary to prevent that amount of energy from rising in temperature above the melting point of the metal (what delta T would that be?)



[/QUOTE]
 
@Chi Meson: k in kT probably is Boltzmann constant.
@hasan_researc:
1/ I think kT only represents the order of magnitude of micro-kinetic energy of particles. I don't really understand the question. Could you write it in an easier-to-understand way?
2/ I'm not sure what kind of light bulb you were mentioning but for a typical incandescent light bulb, it's about 50-100W, and a typical fluorescent light bulb is about 10-40W. You should check your figures.
Another thing to note is that light bulb doesn't convert all electrical energy into radiation. You know, the typical temperature of the filament is around 3000K, so I suppose the temperature of the glass bulb is quite large. That means, there is also heat convection, though rough estimation shows that it's not a big factor.
 
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