A 5.4 g lead bullet moving at 261 m/s strikes a steel plate and stops. If all its kinetic energy is converted to thermal energy and none leaves the bullet, what is its temperature change? Assume the specific heat of lead is 128 J/kg · ◦ C. Answer in units of ◦C.

Change in temperature ∆(tita) is 266.097°C

Explanation:

Ok kinectic energy = 1/2MV²

5.4 grams =( 5.4/1000) kilogram

Kinectic energy =( 1/2 )*(5.4/1000)*261²

Kinectic energy = 183.9267 joules

If kinetic energy = thermal energy

183.9267 joules = mc∆(tita)

Where ∆(tita) = change in temperature

And c = 128 J/kg

∆(tita) = 183.9267/((5.4/1000)*128)

∆(tita) = 266.097

∆(tita) = 266.097°C

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