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Secondary IV • 1yr.

An electronic device consumes 4.1 kW during a week. It dissipates 230 W per day in heat, loses 2000 kW per week due to the light it emits and loses 50 kW per day for another mysterious cause. How much energy is used for the computing functions of the electronic device and what is its percentage of efficiency?

Physics
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Explanations (1)

  • Explanation from Alloprof

    Explanation from Alloprof

    This Explanation was submitted by a member of the Alloprof team.

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    Team Alloprof • 1yr. edited October 2021

    Thanks for your question!

    The gist of your question comes down to the concept of the law of conservation of energy. According to this, in a closed system, the energy will always be the same, since it is never created or destroyed, but simply transformed.

    More precisely, we know the total energy, but we do not know the useful energy of the device:

    • Total energy = 4.1 kW per week

    We know the energy losses of different functions. Sometimes you have to multiply them to get values ​​for the whole week:

    • For heat:

    $$ \frac{230W}{week} • \frac{1kW}{1000W} • \frac {7days}{week} = 1.61kW $$

    • For light (I presume that’s what you mean, otherwise the problem makes no sense;)): 2 kW per week

    • For another mysterious cause (here I presume you mean 0.05 kW and not 50 kW otherwise the problem does not make sense;)):

    $$ 0.05kW • \frac{7 days}{week} = 0.35 kW per week


    We know that the sum of the energy is equal to 4.1 kW. So:

    $$ 4.1kW = 1.61kW + 2kW + 0.35kW + x $$

    where x represents the quantity of useful energy.

    We isolate x and we find that x = 0.14 kW.

    Knowing this, we can answer the second question:

    $$ energy\ efficiency = \frac {energy\ output}{energy\ input} \cdot 100 % $$


    $$ = \frac {0.14kW}{4.1kW} \cdot 100% = 3,41 % $$


    That's it!

    Do not hesitate if you have other questions!