A 4.00-g lead bullet is traveling at a speed of 200 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⋅°C,1 kcal = 4 186 J)
A)78.3 °C
B)157 °C
C)313 °C
D)39.2 °C
E)97.9 °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)(0.004 kg)(200 m/s)^2 = 80 J. Half of this energy remains with the bullet, so 40 J is used to increase its temperature. Convert 40 J to kcal: 40 J / 4186 J/kcal = 0.00956 kcal. The temperature increase is given by Q = mcΔT, where m = 0.004 kg and c = 0.0305 kcal/kg°C. Solving for ΔT gives ΔT = Q/(mc) = 0.00956 kcal / (0.004 kg * 0.0305 kcal/kg°C) = 78.3 °C.

A 400-G Lead Bullet Is Traveling at a Speed of 200