Watts/Heat relationship (dumb question)

Suppose that I have a light bulb consuming 100 watts, a radio also consuming

If I were to measure the heat generated by all these 3 devices, would I get the same amount of heat? In other words, would all devices consuming the same wattage (solid state, no moving parts) generate the same amount of heat?

Thank you.

One more thing, assume they also use the same voltage.

So, does this mean that at the end, they will all be generating the same amount of heat?

The light bulb and the radio will put out a fraction of their energy as light or sound respectively, so it's not all heat. However, it doesn't take long before that light and sound will hit something, not bounce off and be converted into heat.

Tim

Hi,

The light bulb is resistive and is purchased for the value of the watts. This assumes the filament is at the correct temperature for visible light conversion. Other items ( small motors and radios and other items that convert ac into dc are given a rating of peak power ( watts ) so as to be easy to calculate a fuse that should be used in the main switch box to survive repetive re starts after power outages ).

Ray

consuming

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on the lable but uses maybe one half .

Ray

Yes. If you put the devices in light-proof, sound-proof boxes then you would measure that all the boxes would be heated by the same 100W. It'll generally all get converted to heat wherever the light and sound goes, but it's tricky to measure if some of the light gets all the way to Alpha Centauri!

Very strictly speaking some may not end up as heat in certain circumstances - sound or light may end up using some energy breaking chemical bonds for example, but for a radio and light bulb that's insignificant. Even for devices designed to break bonds (like the ultrasound thingies used to smash gallstones) it's a tiny fraction that doesn't end up as heat very quickly.

Tim

Rene,

You are correct. If you measure watts to be equal then you are correct except for the power factor which is ( Power = E x I x Cos of the angle between the voltage and current ) ( theta ) which is different for loads other than resistive. Just measuring the voltage and just measuring the current ( E x I ) gives you apparent power which is always more than that which is creating heat-watts.

Ray

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yes, if they all converted exactly the same amount of energy into something else (movement, light, sound) no, if they all convert different amounts of energy into something else

the bulb will convert an amount into light, the non-light-energy will be turned mostly into heat the heater will have losses in light-radiation (usually IR), the rest will be dissipated as heat the radio will convert energy into sound, the rest will be dissipated as heat

so it all depends on the efficiency of every aparatus that will be compared

Suppose that I have a light bulb consuming 100 watts, a radio also consuming

If I were to measure the heat generated by all these 3 devices, would I get the same amount of heat? In other words, would all devices consuming the same wattage (solid state, no moving parts) generate the same amount of heat?

Thank you.

Peter, With the rising oil/energy prices I may convert back to head phones in the less than 5 watt range.

Ray

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purposes

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Ray

Correct, but as I read the question

- a light bulb and a heater are both resistive loads

- a radio is a more complex load

As I see the question I read "all these apparatus consuming the same wattage..." so I in fact read "do all these apparatus turn the applied (identical) power into heat"

Answer is no, given the fact all of them are designed for different purposes so some of the applied power is turned into something else as intended per design, and the rest into heat

Proof for this can be established empirically

Or just take next situation : I'd like to stand next to my 1kW heater on a cold day, it gets me warm I wouldn't like to stand next to my 1kW amplifier on a cold day, it mostly creates noise but gives me ampel heat :-)

You are correct. If you measure watts to be equal then you are correct except for the power factor which is ( Power = E x I x Cos of the angle between the voltage and current ) ( theta ) which is different for loads other than resistive. Just measuring the voltage and just measuring the current ( E x I ) gives you apparent power which is always more than that which is creating heat-watts.

Ray

something else (movement, light, sound)

consuming

Let's hope you don't actually need 5 watts when using headphones!

There is something called the first law of thermodynamics which states that energy can not be created or destroyed. However, different forms of energy can be interconverted. Thus you can turn electrical energy into heat, light or sound.

However, these energies have different "values" because of the second law of thermodynamics. Crudely speaking, this law states that the amount of disorder ("entropy") increases in an isolated system during each reaction. Thus electrical energy, with its ordered flow of electrons, can be turned into heat (random motion of molecules) with 100% efficiency. The reverse reaction is possible only with a loss, thus the efficiency of a power station is limited to 40% or so.

If the universe is considered an isolated system (which can not be proven), than all forms of energy must eventually be converted to heat because of the second law of thermodynamics, und this heat must distribute evenly throughout the universe. Once this has happend, no further reactions will be possible.

Thus, in this sense, the answer to your question is: Yes.