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Parallel-plate capacitor

  • Thread starter ZEli
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  • #1
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Homework Statement


You need to melt a 0.50 kg block of ice at -10 °C in a hurry. The stove isn't working, but you do have a 50 V battery. It occurs to you that you could build a capacitor from a couple of pieces of sheet metal that are nearby, charge the capacitor with the battery, then discharge it through the block of ice. If you use square sheets spaced 1.8 mm apart, what must the dimensions of the sheets be to accomplish your goal?


Homework Equations


Q(charge) = Aε_0*ΔV/d
Q(heat) = m*L
Q(heat) = mcΔT


The Attempt at a Solution


So far I have calculated the heat required to melt a .5 kg ice cube from -10 °C. The energy required to change the temperature of an ice cube from -10 to 0 I calculated to be
Q = 500g(2.108 J/g*K)(10 K) = 10540. Next, I calculated the heat required to melt the .5 kg ice cube with Q = 500g(334 J/g) = 16700. Thus, the total heat required is 27240 J. However, this is as far as I got. I don't know what else to do! How can I use the energy to find my charge Q?
 

Answers and Replies

  • #2
phinds
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why would you bother bringing a cap into the picture? anything you could do with the cap you could do better with the battery itself.
 
  • #3
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I'm not sure why you're asking me that. I just copy and pasted the question.
 
  • #4
gneill
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I think you slipped a decimal when you melted the ice. Check your heat of melting.

You're going to need to find an expression for the energy stored on a capacitor given the capacitance and voltage. Then you'll need to find the expression for the value of capacitance for a parallel plate capacitor given its dimensions.
 
  • #5
phinds
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I'm not sure why you're asking me that. I just copy and pasted the question.
Fair enough. Book problems do sometimes ask silly questions just so you get a chance to exercise a formula.

Do you understand WHY my question is valid? (I mean about the situation, not about the text-book problem)
 

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