Mechanical equivalent of heat problem

[phhasek]

A 20g lead bullet at 30°C and moving at 350m/s embeds itself in a wodden block.

If 70% of the initial kinetic energy becomes internal energy of the bullet, what is its final temperature?

My attempt:

Ek = (mV^2)/2 = 1225J

70% of this energy = 857.5J

So the available energy to heat the bullet is 857.5J Right?

dQ = m c dt

c - lead = 130J/kg K (specific heat)

dt = (Tf-30°C)

m = 0.02Kg

dQ = 857.5J

Tf = final temperture



So:

857.5 = 0.02 * 130 * (Tf-30)

Tf = 359.8°C

The correct answer should be 327°C

What am I missing?

 

[eljose79]

You are close, but you are using the specific heat of lead instead of the specific heat of the bullet as a whole. Remember, the bullet is not just made of lead, it also has a wooden block surrounding it. So, the specific heat of the bullet will be a combination of the specific heats of both lead and wood. You can calculate this by finding the weighted average of the two specific heats, using the masses of lead and wood in the bullet.

Let's say the bullet is made of 80% lead and 20% wood. This means that 80% of the mass is lead and 20% is wood. So, the specific heat of the bullet will be:

Specific heat of bullet = (0.8 * specific heat of lead) + (0.2 * specific heat of wood)

= (0.8 * 130 J/kg K) + (0.2 * 1700 J/kg K)

= 104 J/kg K + 340 J/kg K

= 444 J/kg K

Now, let's plug this into your equation:

dQ = m c dt

857.5 = 0.02 * 444 * (Tf-30)

Tf = 327°C

So, the final temperature of the bullet after 70% of its kinetic energy is converted to internal energy will be 327°C.