Calculating Melted Ice from a Lead Bullet

[AdnamaLeigh]

Given: latent heat of fusion of water is 333000 J/kg.
A 3.37 g lead bullet at 27 degrees C is fired at a speed of 272 m/s into a large block of ice at 0 degrees C, in which it embeds itself. What quantity of ice melts? Assume the specific heat of lead is 128 J/kg x C. Answer in units of g.

KE = 1/2mv^2
KE = 1/2(.00337)(272)^2 = 124.663 J

KE = Q = mL
124.663 = m(333000) m = .374 g

I don't think that this is correct because I didn't take into account change in temperature of the bullet.

(bullet) mc∆T = mL (ice)

I would set it up like that but I don't have the change in temperature for the bullet. I tried using this formula to figure out the temperature but I got a weird answer:

124.663 = mc∆T = .00337(128)(27 - 0) T= -262 C

That obviously can't be correct. I really need some help.

 

[Fermat]

I think you can assume that ...

1) All the mechanical energy (1/2mv²) is transformed into heat energy used to melt the ice.
2) Everything, the ice, the melted ice/water and the bullet are all at the same temperature, which must be 0 deg C.
3) The mechanical energy plus the heat loss from the bullet are both used to melt the ice.

So, the change in temp of the bullet is 27 deg.

Mind you, I haven't done any thermo in a long time, so I can't guarantee that I'm correct!.

 

[quicknote]

I would suggest that the calculations for this scenario need to take into account the energy transferred from the bullet to the ice, as well as the energy required to melt the ice. The formula you used, Q = mL, only accounts for the energy required to melt the ice (latent heat of fusion), but not the energy transferred from the bullet.

To accurately calculate the quantity of ice melted, we need to use the formula Q = mc∆T, where Q is the energy transferred, m is the mass of the bullet, c is the specific heat of lead, and ∆T is the change in temperature of the bullet.

To find the change in temperature of the bullet, we can use the formula KE = 1/2mv^2, where KE is the kinetic energy of the bullet, m is the mass of the bullet, and v is the velocity of the bullet. This will give us the initial kinetic energy of the bullet before it collides with the ice.

Then, we can use the formula Q = mc∆T to find the energy transferred from the bullet to the ice. We know the mass of the bullet, the specific heat of lead, and we can assume that the final temperature of the bullet and the ice will be 0 degrees C. This will give us the energy transferred from the bullet to the ice.

Finally, we can use the formula Q = mL to find the amount of ice melted, using the energy transferred from the bullet to the ice as the value for Q. This will give us the final answer in units of g.

Overall, it is important to consider all the factors and energy transfers involved in this scenario to accurately calculate the quantity of ice melted.

 

[kyle8921]

I would suggest looking at the problem in a different way. Instead of trying to calculate the change in temperature of the bullet, we can focus on the energy transfer between the bullet and the ice.

First, we can calculate the energy transfer from the bullet to the ice using the formula KE = Q = mL. We know that the kinetic energy of the bullet is 124.663 J, and the latent heat of fusion of water is 333000 J/kg. So, we can calculate the mass of ice melted by dividing the energy transfer by the latent heat of fusion:

m = Q/L = 124.663/333000 = 0.000374 kg = 0.374 g

This means that 0.374 g of ice would melt when the bullet is embedded in the ice.

Next, we need to consider the change in temperature of the ice. Using the formula Q = mc∆T, we can calculate the energy required to raise the temperature of the melted ice from 0°C to its melting point, which is also 0°C:

Q = mc∆T = (0.374)(4186)(0 - 0) = 0 J

This means that no additional energy is needed to raise the temperature of the melted ice to its melting point, as it is already at that temperature.

Overall, we can conclude that 0.374 g of ice would melt from the impact of the lead bullet, assuming that all of the kinetic energy of the bullet is transferred to the ice.