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

  • Thread starter Thread starter ZEli
  • Start date Start date
  • Tags Tags
    Capacitor
AI Thread Summary
To melt a 0.50 kg block of ice at -10 °C using a capacitor charged by a 50 V battery, the total heat required is calculated to be 27,240 J. The discussion emphasizes the need to find the energy stored in a capacitor, which is determined by its capacitance and voltage. The formula for capacitance in a parallel plate capacitor is also relevant, as it relates to the dimensions of the sheets used. There is some confusion about the necessity of using a capacitor instead of directly utilizing the battery's energy. The conversation highlights the importance of understanding the underlying physics in practical applications.
ZEli
Messages
8
Reaction score
0

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?
 
Physics news on Phys.org
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)
 
Thread 'Voltmeter readings for this circuit with switches'
TL;DR Summary: I would like to know the voltmeter readings on the two resistors separately in the picture in the following cases , When one of the keys is closed When both of them are opened (Knowing that the battery has negligible internal resistance) My thoughts for the first case , one of them must be 12 volt while the other is 0 The second case we'll I think both voltmeter readings should be 12 volt since they are both parallel to the battery and they involve the key within what the...
Thread 'Correct statement about a reservoir with an outlet pipe'
The answer to this question is statements (ii) and (iv) are correct. (i) This is FALSE because the speed of water in the tap is greater than speed at the water surface (ii) I don't even understand this statement. What does the "seal" part have to do with water flowing out? Won't the water still flow out through the tap until the tank is empty whether the reservoir is sealed or not? (iii) In my opinion, this statement would be correct. Increasing the gravitational potential energy of the...
Back
Top