Transmission of radiant energy.

Discussion in 'Physics & Math' started by `Juliet., Sep 17, 2009.

  1. `Juliet. Registered Member

    Messages:
    2
    Hello =)

    We carried out an experiement in class, with the aim of determining how effective glass and tinted glass are in cutting down the transmission of radiant energy.

    We were given a formula stating:

    Transmission = Radiant Power transmitted through glass/ radiant power incident on the glass.

    We used a copper calorimeter and a 60W light bulb, as well as the variations of tinted glass.

    Instructions state "Knowing the mass of water, the mass of the calorimeter and the rate of increase in temperature, it is possible to calculate the power incident on the calorimeter. I have this information. I'm assuming that the same information is required to calculate the incident on the tinted glass.

    My question is, how do I carry out the calculations required for this experiment. I really have no idea where to start, and any help would be appreciated.

    Thank you so much!
     
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  3. Pete It's not rocket surgery Registered Senior Member

    Messages:
    10,167
    Use the calorimeter to measure incident power in three situations:

    1. With the light switched off, and no intervening glass.
      This is a control to ensure that the calorimeter is not receiving any other incident power. The measurement should be zero.
    2. With the light switched on, and no intervening glass.
      This is the "radiant power incident on the glass" measurement. (What? But there's no glass in this measurement! Yes, that is correct...)
    3. With the light switched on, and the glass intervening.
      This is the "radiant power transmitted through the glass" measurement.
    Repeat the third measurement for each different kind of glass.
    Repeat the first and second measurements at intervals to check the consistency of the measurement equipment and environment.

    You're welcome!

    Please let me know if I haven't explained it enough, if I've made a mistake, or if I'm just not making sense.
     
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  5. `Juliet. Registered Member

    Messages:
    2

    Hey Pete! Thank you so much for that

    See, I've been able to kindaaaa (not really) get that far, but I'm not sure about what formula I should be using to calculate radiant power incident on the glass. I have results for the changes in temperatureof a period of 10 minutes, taken every 30 seconds. I have the mass of water and the calorimetre. How can i use those to calculate the radient power.

    As you can see, I am absolutely clueless when it comes to this aspect of physics. Sorry!

    thank you so much for your help.
     
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  7. CptBork Valued Senior Member

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    6,465
    Well as Pete's saying, you'd first want to calculate how much energy is being radiated into the calorimeter when there's no intervening glass. The way I would do this experiment, I would have the lightbulb completely immersed inside the calorimeter, and surrounded with some sort of transparent but thermally insultating material, so that most of the energy released is actually in the form of light, as long as the experiment's done over a reasonably short period of time. So with your thermally wrapped light source you know how much light power is being radiated into the calorimeter, and then you can just surround this source with glass and other materials to see how they affect the transmitted radiation.

    There would still be issues to deal with though, such as radiation from the calorimeter leaking back into the original light source (a dark lining on the calorimeter walls would help), and light reflecting around inside the source would still leak out eventually (hopefully you'd have somewhere for it to leak out other than the calorimeter). Not sure what kind of results your class is looking for or how accurately you're looking to demonstrate these phenomena.
     
    Last edited: Sep 21, 2009

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