Yes, I understand. My apologies for not providing a more relevant title.

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    Capacitor
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Homework Help Overview

The problem involves melting a 0.50 kg block of ice at -10 °C using a capacitor charged by a 50 V battery. The original poster calculates the heat required for the phase change and temperature increase but is uncertain about how to relate this to the charge needed from the capacitor.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • The original poster attempts to calculate the total heat required to melt the ice but expresses confusion about how to proceed with using the capacitor. Some participants question the necessity of using a capacitor instead of directly using the battery.

Discussion Status

Participants are exploring different interpretations of the problem, with some suggesting that the original poster may have made an error in their calculations regarding the heat of melting. Guidance has been offered regarding the need to find expressions for energy stored in a capacitor and the capacitance of a parallel plate capacitor.

Contextual Notes

There is a mention of potential errors in the calculations related to the heat of melting, and the discussion touches on the validity of the problem's setup as presented in the homework context.

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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?
 
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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.
 
I'm not sure why you're asking me that. I just copy and pasted the question.
 
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.
 
ZEli said:
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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