Point me in the right direction

[jpnnngtn]

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.

[HallsofIvy]

Originally posted by jpnnngtn
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.

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.

[Meninger]

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.

[Sonty]

And this is college level? During which year of college do you touch QFT, QED, QCD, Group Theory or Representations?

[bogdan]

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...

[Sonty]

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.