14.3b Specific Heat Capacity | A2 Temperature | Cambridge A Level Physics

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  • Опубліковано 27 гру 2024

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  • @mohammedfarhan1565
    @mohammedfarhan1565 3 роки тому +11

    Very helpful keep making these, God bless you all.

  • @tayyabbintahir3197
    @tayyabbintahir3197 Рік тому +8

    Ma'am it would be awesome if you upload a video on the changes in the syllabus according to your playlist(A2)

  • @noxshay2090
    @noxshay2090 Рік тому +3

    U guyz are super amazing I am so thankful respect 3000

  • @legendreflects982
    @legendreflects982 9 місяців тому +1

    we are always gonna assume energy loss is sem for both?

  • @cricketlover22766
    @cricketlover22766 2 роки тому +1

    Miss what is a system and surroundings i mean how to determine what is system and what is surroundings ? In this example is water system and the tube and the rest of the area surroundings?

    • @hammad2536
      @hammad2536 24 дні тому

      System would the things to which you apply energy intentionally and surrounding would be other stuff which you didn't want to get heat up

  • @chinyuandrew7973
    @chinyuandrew7973 5 місяців тому

    30:36

  • @Elpsycongroo1130
    @Elpsycongroo1130 Рік тому

    Energy loss will always be the same for both readings ??

    • @shabananaz9236
      @shabananaz9236 3 місяці тому +1

      It will be same only when for both the different values of rate of water flow(m)--->,(i.e the first and the repeated exp.), we ensure that θ1(inlet temp) and θ2(outlet temp) of water remains unchanged while taking taking these both value of m.
      This is done by adjusting the current I and voltage V during both exp.
      This condition is necessary for 'heat loss' to remain same.

    • @arissmax
      @arissmax 28 днів тому

      @@shabananaz9236 Hi, the emf is constant, do u know why it needs to adjust the variable R to get constant I ? (in my ideal, the i is constant when we do not adjust the R)

    • @shabananaz9236
      @shabananaz9236 28 днів тому +1

      ​@@arissmaxHi, we know that to maintain the same ΔT, power is adjusted. The rheostat is in series with the heater. So consider if we want to increase power(dissipated as heat) across heater, what can we do? To do that, we must decrease rheostat's resistance, so that the total resistance across the circuit decreases as it's= R heater+ R rheostat...this will increase the 'overall current' across circuit (as it's= V total/ R total ..we know😭). The current is same in series everywhere, so now, the 'voltage across the heater' will increase as V heater= I×R heater..(this occurred due increase in I, and ofc heater's resistance is constant).... now you can notice that since 'Power of heater= V heater× I' ..Power will increase(coz of increase in V and I).
      The other way to look at it is by P(heater) =I²× R(heater); the overall increase in current would increase I² value, so that power increases.
      You can look at it the other way to decrease power across heater :)
      About the emf , yea its constant, as it's provided by the source (battery)...

    • @arissmax
      @arissmax 27 днів тому +1

      @shabananaz9236 hi, thanks for ur detailed explanation. I get it now. Ur the best

    • @shabananaz9236
      @shabananaz9236 26 днів тому

      ​@arissmax hi, glad it helped, hehe

  • @learninsane5548
    @learninsane5548 Рік тому +1

    miss ally very beautiful