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)
 
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
Back
Top