# How fast must you throw a snowball against a wall in order to make it melt?

• twotaileddemon
In summary: I'm so sorry, I should have just left it as kJ in the first place.. thanks for all your help though!In summary, the conversation discusses a physics problem involving energy and work. The problem asks how fast a snowball must be thrown against a wall to completely melt, assuming all the energy is transferred to the snowball and air friction is neglected. The conversation explores various equations and concepts related to energy, work, and melting points, ultimately resulting in the solution of the snowball needing to be thrown at a velocity of 25.86 m/s.
twotaileddemon
Hi, I'm new to this forum and I have a question.

In my physics class we are discussing energy and work. We are given this problem:

"It is exactly 0*C. How fast must you throw a snowball against a wall in order to make it completely melt? (Assume all the energy is transferred to the snowball and neglect air friction)"

Now.. considering there is almost no information given, how am I supposed to solve this? I know potential energy is mgh and kinetc is .5mv^2. Work is Fs, or change in kinetic energy. It's just.. give so little details I don't know where to start. Can anyone tell me how to start the problem, or any general guidance without giving the answer? Thanks :)

How much energy do you need to turn ice into water?

I'm not really sure.. we never went over that. Is it just an equation that I'm looking for?

I'm pretty sure kinetic energy is used somehow. And this may be completely off, but could you do something with specific heat or melting/boiling points? I'm given no numerical values to work with.. so I'm a bit lost xx;;

.5m(Vf-Vo)^2 = Ei+El i = initial l = lost
Is this okay?

Yes that's fine. Although I'm not sure what you mean by energy lost, the energy is gained by the snowball and then converted.

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So to solve for Vf, I would get
sq rt (Ei + El + .5mVo^2)/(.5m) = Vf

You need to find the energy needed to melt the snowball. Use the mass and heat of fusion.

$$Ei = 0.5mV_{i}^2 = 0$$
$$mL = 1/2mV_{f}^2$$, L = latent heat of fusion
$$=> V_{f} = sqrt(2L)$$

You can think of m being 1 gram, basically your using a unit mass for L.

edit: put 0 instead of f sorry.

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I guess I'm so confused is because I've never heard the term latent heat of fusion until now, and it was not discussed in class. So sorry if I seem clueless.. because I am xx;;

So the sq rt of mVo^2 would be the correct answer when substituting 1/2mVo^2 for L, ... the 2's cancel out and you're left with sq rt mVo^2? Or do you add the .5mVi^2 to it?

Uh.. oh I see. Vf was already derived. So sqrt of 2L is the answer? There is no other way to simplify it, because my teacher is going to wonder how I came up with that when we didn't discuss it xx;;

I'm so sorry for wasting everyone's time too. I appreciate all the help immensely.

Okay so let me just get something straight. L = 1/2mVf^2
Vf = sqrt 2L

Where does the mass of the snowball go in the equation?
Vf = sqrt 2(1/2mVo^2) = sqrt mVo^2
?

If you read that page you have a value for L, 334.5 joules per gram.
Specific heats and latent heat of fusions are related, see here http://www.rwc.uc.edu/koehler/biophys/8c.html .

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Max Eilerson, why did you choose Ei=0? Snowball has non-zero kinetic energy before collision, zero final kinetic energy.

KEi=(m*v_i^2)/2
This must equal the latent heat of fusion of snowball, which is
L=m*334 kJ/kg

equate these and solve for v_i.
Does that make sense twotaileddemon?
Do you need to know the mass?

Edit: I had typo in specific latent heat value 334

I did read the page, thanks for it.
So Vf is sqrt of 669 J which is about 25.86 J... okay. Thanks :). I will ask my teacher tomorrow whether or not this is allowed since we haven't learned it yet.. maybe he wanted us to do research. Thanks again, I really really appreciate it :)

marcus I got my subscripts confused sorry.

The kinetic energy doesn't equal the latent heat of fusion of the snowball it equals mL.

Max Eilerson, why did you choose Ei=0? Snowball has non-zero kinetic energy before collision, zero final kinetic energy.

KEi=(m*v_i^2)/2
This must equal the latent heat of fusion of snowball, which is
L=m*334 kJ/kg [So do you isolate v_i.. so I get sq rt 334 J / 2kg as my answer? or simplifying it, 117 J / kg? = 117 m/s?

equate these and solve for v_i.
Does that make sense twotaileddemon?
Do you need to know the mass?[No, the velocity it needs to be to melt when it hits a wall]

And.. it makes sense other than the heat of fusion part because I'm just learning it now. .thanks ^_^;

Max Eilerson said:
marcus I got my subscripts confused sorry.

The kinetic energy doesn't equal the latent heat of fusion of the snowball it equals mL.
Sorry, we're using different terms. You call energy/(unit mass) the latent heat, I learned to call it specific latent heat. By that definition
latent heat = specific latent heat * mass
and that gives the equation I supplied.

I don't know which is formally proper, nor which is more commonly used...

They should both work out either way, so as long as you know one you can derive the other by either dividing or multiplying mass ^^

I got 818.5 m/s for the velocity.. I'm pretty sure that's right thanks to both of your help ^^. Thanks so much.

twotaileddemon said:
I got 818.5 m/s for the velocity.. I'm pretty sure that's right thanks to both of your help ^^. Thanks so much.
No your earlier result sqrt(2*334)=25.86 is correct. Check your algebra one more time. Good luck!

marcusl said:
No your earlier result sqrt(2*334)=25.86 is correct. Check your algebra one more time. Good luck!

Actually, I converted kJ to J, so I actually did sqrt (2*334*1000), which was about 818.5.

## 1. How does throwing a snowball make it melt against a wall?

Throwing a snowball against a wall increases its kinetic energy, causing the snow particles to rub against each other and generate heat. This increase in heat causes the snowball to melt against the wall.

## 2. Does the speed of the throw affect how fast the snowball will melt?

Yes, the faster the snowball is thrown, the more kinetic energy it will have upon impact with the wall. This will result in a quicker increase in heat and a faster melting rate.

## 3. Is there a specific speed at which the snowball must be thrown to make it melt against the wall?

No, there is not a specific speed at which the snowball must be thrown to make it melt against the wall. The speed of the throw will affect the melting rate, but other factors such as the temperature of the snowball and the material of the wall will also play a role.

## 4. Can a snowball be thrown too fast for it to melt against the wall?

Technically, yes. If the snowball is thrown at an extremely high speed, it may shatter upon impact with the wall instead of melting. This is because the force of the throw may be too great for the snow particles to withstand, causing them to break apart instead of generating heat.

## 5. Will a snowball melt faster against a warm wall compared to a cold wall?

Yes, a snowball will melt faster against a warm wall compared to a cold wall. This is because the warm wall will provide more heat for the snowball to melt, while a cold wall will absorb the heat from the snowball, slowing down the melting process.

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