Solving a Lightning Storm Homework Problem

vdeity
Messages
6
Reaction score
0

Homework Statement



A certain storm cloud has a potential of 1.50 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 sap in the tree can be boiled away? Model the sap as water initially at 26.0°C. Water has a specific heat of 4186 J/(kg·C°), a boiling point of 100°C, and a latent heat of vaporization of 2.26 x 10^6 J/kg.


Homework Equations



I'm actually kinda iffy on all the equations, but I got several answers that WebAssign told me was wrong...

The Attempt at a Solution



Energy gain of 50.0 C of charge accelerated through potential difference of (1.50 x 10^8 V) is (50.0 C)(1.50 x 10^8 V) = 7.5 x 10^9 J.

To vaporize 1 kilogram of sap requires mcΔT = (1 kg)(4186 J/kg K)(100 - 26 degrees C) = 3.098 x 10^(5) J to heat to 100 degrees C

An additional 2.26 x 10^6 J for the state change from liquid to gas, for a total required energy of 3.098 x 10^(5) + 2.26 x 10^6 = 2.57 x 10^6 J.

The total energy absorbed is 1% of 7.5 x 10^9 = 7.5 x 10^7 J, and at 2.57 x 10^6 J per kg needed, we are able to vaporize:

(7.5 x 10^7 J)/(2.57 x 10^6 J/kg) = 29.19 kg of sap.

WebAssign says it's wrong.

Please tell me where I went wrong.
It isn't due for a few days, I just wanted to get this assignment out of the way... Last question.

Thanks!
 
Physics news on Phys.org


WebAssign is evil. I see nothing wrong with your solution - your method is OK and I verified your numbers. It may be a matter of sig figs or (perish the thought) the wrong calculated formula has been put into the problem. I suggest that you take your solution to your instructor and ask him/her to find what is wrong with it. Do this before the deadline.
 


Sorry to bring up an old thread, but I was under the impression that the energy absorbed by the tree will be equal to 1/2QV, and not QV. I'd be curious to hear what others have to say about this.

Thanks,

Tom
 


That would be true if you had a capacitor with 50C on it that you let completely discharge.
The question doesn't say that the stormcloud completely discharges however.
 


I see what you're saying. Thanks.
 
Hi, I had an exam and I completely messed up a problem. Especially one part which was necessary for the rest of the problem. Basically, I have a wormhole metric: $$(ds)^2 = -(dt)^2 + (dr)^2 + (r^2 + b^2)( (d\theta)^2 + sin^2 \theta (d\phi)^2 )$$ Where ##b=1## with an orbit only in the equatorial plane. We also know from the question that the orbit must satisfy this relationship: $$\varepsilon = \frac{1}{2} (\frac{dr}{d\tau})^2 + V_{eff}(r)$$ Ultimately, I was tasked to find the initial...
The value of H equals ## 10^{3}## in natural units, According to : https://en.wikipedia.org/wiki/Natural_units, ## t \sim 10^{-21} sec = 10^{21} Hz ##, and since ## \text{GeV} \sim 10^{24} \text{Hz } ##, ## GeV \sim 10^{24} \times 10^{-21} = 10^3 ## in natural units. So is this conversion correct? Also in the above formula, can I convert H to that natural units , since it’s a constant, while keeping k in Hz ?
Back
Top