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

  • Thread starter ZEli
  • Start date
  • Tags
    Capacitor
In summary, the conversation discusses the process of melting a 0.50 kg block of ice at -10 °C using a 50 V battery and a capacitor made from sheet metal. The individual has calculated the heat required to melt the ice but is unsure how to use this information to find the charge Q. They are also unsure of the purpose of using a capacitor in this situation. The conversation concludes with a discussion on finding the energy stored on a capacitor and the expression for capacitance for a parallel plate capacitor.
  • #1
ZEli
8
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
  • #2
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
I'm not sure why you're asking me that. I just copy and pasted the question.
 
  • #4
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
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)
 

1. What is a parallel-plate capacitor?

A parallel-plate capacitor is a type of capacitor that consists of two parallel conductive plates separated by a dielectric material. It is used to store and release electrical energy in electronic circuits.

2. How does a parallel-plate capacitor work?

When a voltage is applied across the plates, an electric field is created between them. The electric field causes a buildup of charges on the plates, with positive charges on one plate and negative charges on the other. This separation of charges creates a potential difference, or voltage, between the plates.

3. What is the capacitance of a parallel-plate capacitor?

The capacitance of a parallel-plate capacitor is determined by the size of the plates, the distance between them, and the type of dielectric material used. It is directly proportional to the area of the plates and inversely proportional to the distance between them.

4. How is a parallel-plate capacitor different from other types of capacitors?

A parallel-plate capacitor has a simple and symmetric design, making it easier to calculate and control its capacitance. It also has a higher capacitance compared to other types of capacitors with the same size and distance between plates. However, it is not suitable for high voltage applications due to the risk of arcing between the plates.

5. What are some common uses of a parallel-plate capacitor?

Parallel-plate capacitors are commonly used in electronic circuits to filter out high-frequency noise, store energy in power supplies, and tune radio frequencies. They are also used in sensors, accelerometers, and microelectromechanical systems (MEMS).

Similar threads

  • Introductory Physics Homework Help
Replies
3
Views
983
  • Introductory Physics Homework Help
Replies
10
Views
2K
  • Introductory Physics Homework Help
Replies
8
Views
2K
  • Introductory Physics Homework Help
Replies
4
Views
2K
  • Introductory Physics Homework Help
Replies
3
Views
1K
  • Introductory Physics Homework Help
Replies
7
Views
2K
  • Introductory Physics Homework Help
Replies
3
Views
1K
  • Introductory Physics Homework Help
Replies
5
Views
5K
  • Introductory Physics Homework Help
Replies
2
Views
3K
  • Introductory Physics Homework Help
Replies
4
Views
944
Back
Top