Melting a snowball by throwing it at a wall

In summary, the conversation discusses how to calculate the speed at which a snowball must be thrown in order to completely melt upon impact with a brick wall. The suggested approach is to use the equations Q = mL (where Q represents energy, m represents mass, and L represents the latent heat of fusion) and E_k = 1/2 * mv^2 (where E_k represents kinetic energy and v represents velocity). After some discussion about the need for mass in the equation, it is determined that the mass can be assumed to be 1 kg, resulting in a final calculation of v = √(2L).
  • #1
AbsoluteZer0
125
1

Homework Statement



You throw a snowball at 0.0 Celsius at a brick wall. If you want it to melt completely, how fast will you have to throw it?

Homework Equations



[itex] Q = mL [/itex]

[itex] E_k = \frac{1}{2} mv^2 [/itex]

The Attempt at a Solution



I initially reasoned that you would use Q = mL to find the energy needed to melt it and then substitute that into the Kinetic energy formula, giving:

[itex] mL = \frac{1}{2} mv^2 [/itex]

However, I am given no information regarding the mass of the snowball and haven't figured out a method to find that mass.
Any suggestions?

Thanks,
 
Physics news on Phys.org
  • #2
Why do you need mass when it cancels out in the equation you formed? :confused:
 
  • #3
AbsoluteZer0 said:

Homework Statement



You throw a snowball at 0.0 Celsius at a brick wall. If you want it to melt completely, how fast will you have to throw it?

Homework Equations



[itex] Q = mL [/itex]

[itex] E_k = \frac{1}{2} mv^2 [/itex]

The Attempt at a Solution



I initially reasoned that you would use Q = mL to find the energy needed to melt it and then substitute that into the Kinetic energy formula, giving:

[itex] mL = \frac{1}{2} mv^2 [/itex]

However, I am given no information regarding the mass of the snowball and haven't figured out a method to find that mass.
Any suggestions?

Thanks,

Just assume that mass equals 1.
 
  • #4
Pranav-Arora said:
Why do you need mass when it cancels out in the equation you formed? :confused:

Ah!
I didn't notice that. Thanks.

I'm now at

[itex]L = \frac{1}{2}v^2 [/itex]

Edit:

I changed 334 j/g to 334000 j/kg and arrived at around 800 m/s
 
Last edited:
  • #5
If you want your working checked, please post it.
 

FAQ: Melting a snowball by throwing it at a wall

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

Throwing a snowball at a wall causes it to melt due to the impact and friction between the snowball and the wall. This creates heat, which causes the snow to melt.

2. Why does a snowball melt when it hits a wall?

When a snowball hits a wall, the kinetic energy from the impact is converted into heat energy. This heat energy increases the temperature of the snowball, causing it to melt.

3. Can any type of wall melt a snowball?

No, not all walls can melt a snowball. The wall must be made of a material that can conduct heat, such as concrete or brick, in order for the snowball to melt upon impact.

4. How long does it take for a snowball to melt when thrown at a wall?

The time it takes for a snowball to melt when thrown at a wall depends on various factors such as the size and density of the snowball, the speed and force of the throw, and the material and temperature of the wall. Generally, it can take a few seconds to a few minutes for a snowball to melt upon impact.

5. Can a snowball melt completely when thrown at a wall?

It is possible for a snowball to melt completely when thrown at a wall, but it depends on the factors mentioned earlier. If the snowball is large and dense, and the wall is warm and made of a good conductor of heat, then it is likely that the snowball will melt completely. However, if the snowball is small and the wall is cold, the snowball may not completely melt upon impact.

Back
Top