Question

A cowboy fires a silver bullet with a muzzle speed of 200 m/s into the pine wall of a saloon. Assume all the internal energy generated by the impact remains with the bullet. What is the temperature change of the bullet?

Answer 1

Explanation:

KE = q

½ mv² = mCΔT

ΔT = v² / (2C)

ΔT = (200 m/s)² / (2 × 236 J/kg/°C)

ΔT = 84.7°C

Answer 2

This question involves the concepts of the law of conservation of energy.

The temperature change of the bullet is "84.38°C".

What is the Law of Conservation of Energy?

According to the law of conservation of energy, total energy of the system must remain constant. Therefore, in this situation.

$Kinetic\ energy\ of\ bullet\ before\ impact=heat\ absorbed\ in\ bullet\\\\\frac{1}{2}mv^2=mC\Delta T\\\\\Delta T = \frac{v^2}{2C}$

where,

• ΔT = change in temperature of the bullet = ?
• C = specific heat capacity of silver = 237 J/kg°C
• v = speed of bullet = 200 m/s

Therefore,

$\Delta T = \frac{(200\ m/s)^2}{2(237\ J/kg.^oC)}$

ΔT = 84.38°C

Learn more about the law of conservation of energy here:

https://brainly.com/question/20971995

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