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

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SUMMARY

The discussion focuses on estimating the mass of train brakes required to prevent overheating and the number of photons emitted by a light bulb. The user calculated the kinetic energy of a train moving at 30 m/s with a mass of 2000 kg, resulting in an absurd temperature of 3.2 x 1030 K, indicating a misunderstanding of energy versus temperature. The correct approach involves calculating the heat energy (Q) and determining the mass needed to keep the temperature below the melting point of the brake material. For the light bulb, the power was calculated at 380 mW, leading to an estimate of 1.3 x 1012 photons emitted per second.

PREREQUISITES
  • Understanding of kinetic energy calculations
  • Familiarity with thermal energy concepts
  • Knowledge of photon energy calculations
  • Basic principles of heat transfer and material melting points
NEXT STEPS
  • Research the relationship between kinetic energy and temperature in materials
  • Learn about the Boltzmann constant and its applications in thermal physics
  • Explore the efficiency of different types of light bulbs and their thermal properties
  • Investigate methods for calculating mass and heat transfer in braking systems
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Students in physics or engineering, particularly those studying thermodynamics and energy transfer, as well as professionals involved in mechanical design and safety analysis of braking systems.

hasan_researc
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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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