A 4.6 g lead bullet moving at 278 m/s strikes a steel plate and stops. If all its kinetic energy is converted to thermal energy

Question

3 years ago

5

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.

1 answer:

iren2701 [21] · Answer 1 · 2022-06-30T23:02:59+03:00

Answer:

ΔT = 302 °c

Explanation:

mass (m) = 4.6 g = 0.0046 kg

velocity (v) = 278 m/s

specific heat of lead (c) = 128 J/kg. °c

kinetic energy = 0.5 mx $v^{2}$

kinetic energy = 0.5 x 0.0046 x $278^{2}$

kinetic energy = 177.8 J

since all the kinetic energy is converted to thermal energy,

kinetic energy = thermal energy (E) = 177.8 J

thermal energy = m x c x ΔT

where ΔT is the temperature change

177.8 = 0.0046 x 128 x ΔT

ΔT = 177.8 / 0.59

ΔT = 302 °c