# Homework Help: Heat Energy and Change in Temperatur Problem - Bullet Kinetic Energy into Heat Energy

1. Jun 15, 2010

### leslie123

1. The problem statement, all variables and given/known data
A 5 g lead bullet moving at 343 m/s is stopped in a block of wood. All of the kinetic energy goes into heat energy added to the bullet. The initial temperature of the bullet is 24 degrees C. What is the final temperature of the bullet in degrees C? (cPb = 0.128 kJ/kg . K, LfPb = 24.7 kJ/kg, MPPb = 600 K)

2. Relevant equations
K = 1/2*m*v^2
Q = mc$$\Delta$$T

3. The attempt at a solution
1/2mv$$^{2}$$=mc$$\Delta$$$$T$$
1/2v$$^{2}$$ = c$$\Delta$$T

Using this equation $$\Delta$$T = 460C

I realise that the mass must come into it with the other provided information but can not work out how.

Last edited: Jun 15, 2010
2. Jun 15, 2010

### RoyalCat

Re: Heat Energy and Change in Temperatur Problem - Bullet Kinetic Energy into Heat En

When approaching such a problem, one of the most important things to remember is phase transitions. When you heat a substance past its melting point, whatever energy you add to it, goes into melting another portion of the mass (The concept of latent heat is useful in solving such problems, note that it was given to you.)

Another useful thing to remember is to never mix your units, having some data in degrees Celsius, and the rest in kelvins is confusing, move everything to kelvins and see what you can say about what's happening.

3. Jun 15, 2010

### mikelepore

Re: Heat Energy and Change in Temperatur Problem - Bullet Kinetic Energy into Heat En

You said the mass must come into it. But why? In your step 3 you showed that "m" cancels out.

4. Jun 15, 2010

### leslie123

Re: Heat Energy and Change in Temperatur Problem - Bullet Kinetic Energy into Heat En

Ok, once I started working in the same units (Thanks!) I realised that with a temperature change of 460K, the bullet has easily reached its melting point.

As the bullet starts at 297K it must use 303K worth of heat to reach its melting point.
Using Q=mc$$\Delta$$T plugging in Q = (5*10^-3)(0.128*10^3)(303) we have 194J to heat the bullet up to melting point.

Using Q = mL plugging in Q = (5*10-3)(24.7*10^3) the system needs 124J to fully melt the bullet.

Using Q=mc$$\Delta$$T plugging in Q = (5*10^-3)(0.128*10^3)(460) we have 294J total energy going into the system.

As the bullet will use all of the energy up (194J to heat it up and 100J to melt it) before it is fully melted is this the reason that it stays at 600K (327C)? If the bullet had enough energy to melt it would it continue to heat up?

5. Jun 15, 2010

### RoyalCat

Re: Heat Energy and Change in Temperatur Problem - Bullet Kinetic Energy into Heat En

The question is whether there is enough energy to melt the entire bullet, or not. If there's only enough energy to melt some of the lead, then the system will end up at the melting point of lead, since it can't go over that temperature until all of the lead has melted.

Once the entire bullet has melted, whatever energy you put into it goes into warming it up. I think the assumption you're expected to work under here is that the heat capacity of the molten lead is the same as that of the solid lead, which may not be too realistic, but it's what we've got. :)

Looking it up on the internet I found that the molar specific heat capacity of liquid lead is $$c=28.6 \frac{J}{mol \cdot K}$$. Comparing that with the value for solid lead, which is $$26.6 \frac{J}{mol \cdot K}$$ justifies our assumption.

6. Jun 15, 2010

### leslie123

Re: Heat Energy and Change in Temperatur Problem - Bullet Kinetic Energy into Heat En

Awesome, thanks!