I tried using:
.7E = 1/2mv^2 - cm(dT)
However, I do not know the total energy of the system, so I can't finish the problem this way.
A 19-g lead bullet is shot with a speed of 340 m/s into a wooden wall. Assuming that 70% of the kinetic energy is absorbed by the bullet as heat (and 30% by the wall), what is the final temperature of the bullet? (Assume the bullet is initially at room temperature of 20°C. Use the numerical data found in this table.)
I know the answer is 334 degree C. How do I get this answer?
3 answers
The table gives the specific heat of lead as 0.129 KJ/(kgK) and 0.0308 cal/(gK).
0.7•KE=Q
0.7•mv²/2=mc(tº-20º)
tº= 20º+0.7•v²/2 =
=20º+0.7•340²/2•129= 333.6ºC≈334ºC
0.7•mv²/2=mc(tº-20º)
tº= 20º+0.7•v²/2 =
=20º+0.7•340²/2•129= 333.6ºC≈334ºC
1 answer