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A 5.0 g lead bullet is fired into a fence post. The initial speed of the bullet is 210 m/s, and when it comes to rest, half its kinetic energy goes into heating the bullet. How much does the bullet's temperature increase?

Verified Answer :

MonsieurPengu

5g = 0.005 kg
So KE=1/2 0.005 kg 210m/s^2 = 110 J.
Now the specific heat capacity of lead is 126 J/kgK, so the temperature change= 55 J / 0.005 126 J/kgK=87.3015873 K. Which, if normal thermal conditions, the bullet would have a temperature of around 360,451587 Kelvin or, simplified, 107.3 degrees celsius.

Eduard22sly

Given the data from the question, the bullet's temperature increases by 85.5 °C

How to determine the kinetic energy

  • Mass (m) = 5 g = 5 / 1000 = 0.005 g
  • Velocity (v) = 210 m/s
  • Kinetic energy (KE) =?

KE = ½mv²

KE = ½ × 0.005 × 210²

KE = 110.25 J

How to determine the increase in temperature

  • Kinetic energy (KE) = 110.25 J
  • Heat (Q) = ½KE = ½ × 110.25 = 55.125 J
  • Mass (M) = 5 g
  • Specific heat capacity of lead (C) = 0.129 J/gºC
  • Change in temperature (ΔT) =?

Q = MCΔT

Divide both side by MC

ΔT = Q / MC

ΔT = 55.125 / (5 × 0.129)

ΔT = 85.5 °C

Thus, the temperature increase is 85.5 °C

Learn more about heat transfer:

https://brainly.com/question/6363778

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