A 3.00 g lead bullet is traveling at a speed of 300 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 = 0.0305 kcal/kg·$\degree$C, 1 kcal = 4186 J)
A) 113$\degree$C
B) 137$\degree$C
C) 176$\degree$C
D) 259$\degree$C

Question

Quizplus · Accepted Answer

The correct answer is C.The heat energy generated is:$$Q = \frac{1}{2}mv^2 = \frac{1}{2}(0.003 kg)(300 m/s)^2 = 135 J$$The increase in temperature is:$$\Delta T = \frac{Q}{mc} = \frac{135 J}{(0.003 kg)(0.0305 kcal/kg·°C)(4186 J/kcal)} = 176 °C$$

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