Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Point me in the right direction

  1. Aug 3, 2003 #1
    Here's the problem:

    A certain storm cloud has a potential difference of 1.00 X 10^8 V relative to a tree. If, during a lightning storm, 50.0 C of charge is transferred through this potential difference and 1.00% of the energy is absorbed by the tree, how much water (sap in the tree) initially at 30 degrees Celsius can be boiled away? Water has a specific heat of 4186 j /kg * degrees Celsius. It has a boiling point of 100 degrees Celsius, and a heat of vaporization of 2.26 X 10^6 J /kg

    I know that this problem has to do with Energy stored in a capacitor.

    Potential Difference = 1.00 X 10^8

    C = e(o)(A / d) "I dont have an area or a distance so that wouldn't matter"

    I think what is throwing me off is the temperature included in this problem, for that matter, everything involving the water.
  2. jcsd
  3. Aug 4, 2003 #2


    User Avatar
    Science Advisor

    I don't think it is the temperature so much as the "area-distance" problem. It's interesting that you give a formula that involves area and distance and then say "I dont have an area or a distance so that wouldn't matter"! What you really mean is that that formula is irrelevant.

    The crucial point is that a "volt" is a measure of potential difference per coulomb. Since you are told that the lightening had a voltage of 1.00 X 10^8 V and the tree was hit by 50 Coulombs, there was an energy increase of 50 X 10^8 Joules.

    Now, use the temperature information.

    You know that the specific heat of water is 4186 j /kg * degrees Celsius so it takes 70* 4186= 293020 J to raise each kg of water from 30 to 100 degrees C. That is, if the mass of water is M, The energy required to raise it to 100 degrees C is 293020*M Joules which is
    2.93 X 10^5*M Joules (3 significant figures since heat of vaporization is only given to 3 significant figures).
    You also know that the "heat of vaporization" for water is 2.26 X 10^6 J /kg so the amount of energy necessary to vaporize M kg. of water is 2.25 X 10^6 *M= 22.5 X 10^5 *M Joules.

    To raise M kg of water from 30 degrees to 100 degrees and then vaporize it requires: 2.93 X 10^5*M+ 22.5 X 10^5*M =
    25.4 X 10^5*M Joules. Since you know the total Joules available is
    50 X 10^8 Joules. Solve for M.
  4. Aug 5, 2003 #3
    Remember that electrical energy equals Vq

    You know V and you know q

    solve for EPE

    find one percent of this

    Look in a first semester physics book to find equations you can plug in to solve for the rest.
  5. Aug 10, 2003 #4
    And this is college level? During which year of college do you touch QFT, QED, QCD, Group Theory or Representations?
  6. Aug 10, 2003 #5
    It depends on what college it is...because they learn many different things...not necessarily at the highest level...
    I saw a paper about artificial neural nets for students who were studying something completely different...
  7. Aug 11, 2003 #6
    This is much to juicy to discuss here. I'll open a thread about what classes does each of us take.

    P.S. It's funny to see two romanians talking in english.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook