A 3.00-g lead bullet is traveling at a speed of 240 m/s when it embeds in a wood post.If we assume that half of the resultant heat energy generated remains with the bullet,what is the increase in temperature of the embedded bullet? (specific heat of lead = 0.030 5 kcal/kg* $\degree$C,1 kcal = 4 186 J)
A)113 $\degree$C
B)137 $\degree$C
C)226 $\degree$C
D)259 $\degree$C

Question

Quizplus · Accepted Answer

The kinetic energy of the bullet is converted into heat energy. The kinetic energy is (1/2)mv^2 = (1/2)(0.003 kg)(240 m/s)^2 = 86.4 J. Half of this energy remains with the bullet, so 43.2 J is used to increase its temperature. Using the specific heat formula Q = mcΔT, where Q = 43.2 J, m = 0.003 kg, and c = 0.0305 kcal/kg°C = 0.0305 * 4186 J/kg°C, we solve for ΔT to find the increase in temperature.

A 300-G Lead Bullet Is Traveling at a Speed of 240