How much will a lead bullet's temperature rise if it stops without losing heat?

[Karol]

Homework Statement

A lead bullet flies at 350[m/s], hits and stops. if no heat escapes, how much will it's temperature rise

Homework Equations

$Q[cal]=m\cdot c\cdot \Delta t$
Specific heat of lead: c=0.031
1[cal]=4.186[Joule]

The Attempt at a Solution

The kinetic energy: $E=\frac{1}{2}mv^2=\frac{1}{2}m\cdot 350^2=m\cdot 61,250[Joule]$
$\frac{61,250}{4.186}=14,632[cal]\rightarrow E=m\cdot 14,632[cal]$
$Q=E=m\cdot 14,632[cal]=m\cdot 0.031\cdot \Delta T$
And it gets a huge temperature

 

[ehild]

The energy is in joules if m is in kg-s. At the same time you use cal/(g C°) for specific heat. So the m-s on both sides of the last equation are not the same, it is in kg-s on the left side and grams on the right side. Better to use the SI units.

ehild

 

[Karol]

The kinetic energy: $E=\frac{1}{2}mv^2=\frac{1}{2}m\cdot 350^2=m\cdot 61,250[Joule]$
$\frac{61,250}{4,186}=14.6[kcal]\rightarrow E=m\cdot 14.6[kcal]$
The specific heat doesn't change it's value, 0.031, because of the unit change:
$Q=E=m\cdot 14.6[kcal]=m\cdot 0.031\left[\frac{kcal}{kg\cdot k^0}\right] \cdot \Delta T\rightarrow Delta T=472^0$
Is that logical? the melting point of lead is 327⁰

 

[NTW]

May I make a suggestion? One should never use, in decimal fractions, three figures. Thus, the risk of confusion is easily avoided.

For example, when converting to SI units the sp. heat of lead, I would write 0,0310 cal/g*ºC x 4,1860 Joule/cal = 0,1298 J/g*ºC = 129,8 J/kg*K

A comment: I went to school in a 'decimal point country', and later moved to a 'decimal comma country'... I have always avoided three decimals, and even use the elevated comma to write them... 129'8 is, for example, my usual way of writing decimals... Not unusual in Spain, even if this is a 'decimal comma country'...

I see that the problem has been solved. So, I believe that I can now write down my solution, using SI units throughout:

Delta T = Energy of the bullet/specific heat * mass

As the kinetic energy of a bullet of m mass at 350 m/s is 61250m Joule, we have:

Delta T = 61250m Joule/((129,8 J/Kg*K )* m) = 61250/129,8 = 472 K of rise in temperature...

 

[Karol]

The kinetic energy: $E=\frac{1}{2}mv^2=\frac{1}{2}m\cdot 350^2=m\cdot 61250[Joule]$
$Q=E=m\cdot 61250[kcal]=m\cdot 129.8\left[\frac{Joule}{kg\cdot k^0}\right] \cdot \Delta T\rightarrow Delta T=472^0$
The same high result, does the bullet melt??

 

[NTW]

That's not relevant. This is a physics problem... In real life, it is difficult that the whole of the kinetic energy resulting from the strike may be converted in heating the bullet, as the target is usually deformed permanently and heats up somewhat, the bullet may permanently deform or break, (something that needs energy) it may recoil, or the pieces may recoil, taking away part of the kinetic energy, and that may result in other permanent deformations elsewhere, and in an increase of the temperature of the surroundings...

 

[billy_joule]

Is that logical? the melting point of lead is 327⁰

I think it's logical.
For a real bullet most of the energy is spent blowing someone's skull apart Which probably takes a fair amount of energy, which if converted to heat would be significant.

It's always good to check whether your answer makes some sort of physical sense. But in this case, as NTW points out it's hard to do.

 

[ehild]

Do not expect a problem maker to think of everything... :) The bullet stops in such a short time that it can not melt yet. It gets hot, it starts to melt and starts to give heat to the surroundings.
By the way, higher speed bullets are made of some alloy, which melts at higher temperature.

ehild

 

[Karol]

Thank you everybody, i won't mess with bullets...

 

[zoki85]

Is that logical? the melting point of lead is 327⁰

The bullet gets heat, but target gets heat too