Connecting Electric Potential and Thermal Energy

[RIPCLB]

Homework Statement

The question states:

You notice a thuderstorm and calculate the Potential Difference between a cloud and a tree to be 150 MV. You know that a lightning bolt delivers 60 C of charge. If the tree only absorbs 5% of the energy of this lightning bolt, with the rest going to the ground, if they tree is at 30 degrees C, how much water can be boiled away within the tree? Water has a specific heat of 4186 K/Kg degrees C, and boiling point is 100 degrees C, heat of vaporization is 2.26 x 10^6 J/Kg

Homework Equations

V = PE/q
Q=MC delta T
Q = ML

There must be something with Density, but I'm not completely sure.

The Attempt at a Solution

I figured out using V = PE/q (Solving for PE) that the Potential Electric Energy is 9 x 10^9, and that 5% of that is 45 x 10^7 (I'm pretty sure that's correct, but if I'm wrong let me know). From there, I'm not all to sure where to go since I'm not given an initial mass or a final temperature. Any help would be greatly appreciated.

 

[lanedance]

Hi

could make the assumption all of the 5% goes towards boiling water (and only that) & use that info to work out a mass...

 

[RIPCLB]

I do know that the answer is 88.1 kg (we're given the answer so we can work out the process), which is something I forgot to add earlier.

 

[lanedance]

first work out how much energy it takes to boil 1kg, this is heating the water 70degC the vaporising it

then divide the 5% of electrical energy by this value & it will give you the mass of water you can boil

equivalent to writing
0.05*PE = M*(c.dT + L) and solving for M

 

[RIPCLB]

Thats what I had thought to do, but the number ends up being too high, maybe I have the wrong PE?

 

[lanedance]

yeah seems to be twice as big, so maybe we assume the potential difference between the cloud & gorund is zero after the strike, this means V vareis linearly from 150MV to zero as the charge is is treansferred giving,

then the energy is W = \intv.dq

with v(q) = V0.(q-q0)/q0
and integrate from 0 to q0

which gives PE = V0.q0/2

so effectively the cloud ground system is working like a capacitor with Capacitance q0/V0