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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)

Respuesta :

Answer:

 ΔT = 78.32 °C

Explanation:

Kinetic energy of bullet is K

         K = mv²/2

            = .004 * ( 200 m/s )²/2

            = 80 J

Half of the heat E = 40 J

Now a per the equation of energy transferred ,

                        E = cm ΔT

c is the specific heat of latent, m is the mass and ΔT is the change in temperature.

                      ΔT = 40 / ( .0305 × 4186 × .004)

                      ΔT = 78.32 °C

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