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August 27, 2014

August 27, 2014

Posted by **Ceres** on Wednesday, September 23, 2009 at 4:07pm.

"A block of ice at 0oC is dropped from a height that causes it to completely melt upon impact. Assume that there is no air resistance and that all the energy goes into melting the ice. Show that the height necessary for this to occur is at least 34 km."

I know that it takes 334J/g to turn ice to liquid (thanks to help that was given to me for another problem). Does this have anything to do with this problem? no mass is given for the ice, so I assume that there is no difference for one gram than there is for many gms of ice. What does the height have to do with the ice turning to water? The higher up it drops from the more force it has? Thanks.

- Physics -
**MathMate**, Wednesday, September 23, 2009 at 7:11pm"I know that it takes 334J/g to turn ice to liquid "

That's a good start.

The question assumes no air-resistance, which means that the potential energy at the high altitude instantly and completely transforms into heat energy and distributed evenly to the piece of ice to melt it. It further assumes that the value of g, acceleration due to gravity does not change with elevation (height).

Potential energy when an object of mass m is positioned at a height h from ground is mgh, where g is the acceleration due to gravity.

So equating heat energy with potential energy, consider a 1-kg block of ice:

334 joules/g *1000g = 1 kg*9.8*h

Find h.

- Physics -
**Ceres**, Wednesday, September 23, 2009 at 9:12pmI am trying to understand what you wrote. Where did the 1000g come from? And how do we know the mass of the ice is 1kg? There is no weight for the ice listed in the problem. I did solve for h and got 34.1. I also didn't realize that heat energy would be the same as potential energy, so thank you for pointing that out. Sorry to be a bother, but I really want to understand why the problem works as it does. Thanks!

- Physics -
**MathMate**, Wednesday, September 23, 2009 at 10:01pmIn fact, the 1 kg is not necessary because in the equation, the mass will cancel out on each side. I used (without elaboration) 1 kg just for illustration purposes. My apologies.

- Physics -
**Ceres**, Thursday, September 24, 2009 at 12:43amThank you for your clarification.:-) I wasn't sure if you would notice a post so far back - I'm new to this site.

I am sorry to bother you again. I can post this as a new question if you would rather. But the next question is along the same lies, except this time it is a iron ball that drops and not ice. It also uses the specific heat capacity of the iron. The questions is:

A 10-kg iron ball is dropped onto a pavement from a height of 100 m. Suppose half of the heat generated goes into warming the ball. Show that the temperature increase of the ball is 1.1oC. (In SI units, the specific heat capacity of iron is 450 kJ/kg*oC.) Why is the answer the same for an iron ball of any mass?

From the last question I know that the equation will have mass on both sides, just like the ice problem. But no water is present so I can't use the heat of fusion since the iron does not melt (although it does warm). Can you help me understand why and how I would use the heat capacity of iron to solve this problem? I have been working on the same five problems from this chapter all day and am really frustrated. Thanks for your help.

- Physics -
**MathMate**, Thursday, September 24, 2009 at 6:53amThe basic principle is the conservation of energy.

When a ball is brought higher up by h=100 m., energy is required to make this happen. The energy is stored as potential energy, Ep=mgh, m=mass, g=acceleration due to gravity.

When the ball falls from this height in a free fall, energy is transformed into kinetic (movement) energy, given by the equation Ek=(1/2)mv², where v=velocity of the ball.

When the ball hits the ground, the velocity is reduced, so is the kinetic energy according to the formula Ek. Since total energy is conserved, the difference goes into other forms, such as kinetic energy of the fragments of soil flying around (sum of (1/2)mv² of the soil fragments), and the rest (in this case half) goes into heat, given by

Eh=mH, H=specific heat, and m again the mass.

I recapitulate:

The basic principle is the conservation of energy, in different forms.

Hope this clears up a little more of the picture. Post any time for more explanations.

- Physics-corr -
**MathMate**, Thursday, September 24, 2009 at 7:27amCorrection:

EhΔT=mH, H=specific heat, m again the mass, and ΔT=change in temperature.

Also, you have done the right thing in posting in both places, the original post and a new- or re-post as a reminder. As Sara said, a reference to the original post (Date and time, or the 10-digit id of the original post) would have helped other teachers to go back to the original context of the question. Did you have to retype the question and the responses? In fact, it so happened that I have checked the original post before the new ones, but it was more an exception than anything else. I apologize for the comment.

- Physics-corr. -
**MathMate**, Thursday, September 24, 2009 at 7:28amThe ΔT was in the wrong place, sorry.

Eh=mHΔT, H=specific heat, m again the mass, and ΔT=change in temperature.

- Physics -
**Ceres**, Thursday, September 24, 2009 at 12:10pmI worked on this the rest of last night and some more this morning and this is what I got.

Ep = mgh

Ep = 10kg(9.8m/s2)(100m)

Ep = 9800kgm2

I then equated that to the kenetic energy and used that amount to find the velocity:

9,800kg(m2) = 1/2 mv2

square root of 1960 = v

Now that I'm thinking it through more, I'm not sure why I did the calculations above.

Next I plugged in the numbers to find the change in temperature. I got:

Eh = mH(chng in Temp)

Eh = 10kg(450kj/kg*oC)(chng in T)

But I do not know the Eh (I'm not sure what that stands for) nor am I supposed to use the 1.1oC for the change in temp.

I feel as if I am hopelessly dense. I understand all of the concepts you talked about - conservation of energy, potential energy to kenetic energy, etc. But my brain is still missing something and I'm not sure what it is. I included the above equations because maybe they would help you to see where my thinking has gone wrong. Thanks again for your help. I will copy and repost this because it has been so long since I last posted.

- Physics -
**Mlungisi**, Tuesday, June 21, 2011 at 9:47am34,08km

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