A 55.0-gram lead bullet hits a steel plate, both initially at 24 °C. The bullet melts and splatters on impact. Assuming that the bullet receives 72% of its kinetic energy as heat energy, at what minimum speed must it be traveling to melt on impact? (The melting temperature of Lead is 327°C)

Question

25-11-2019
Physics

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A 55.0-gram lead bullet hits a steel plate, both initially at 24 °C. The bullet melts and splatters on impact. Assuming that the bullet receives 72% of its kinetic energy as heat energy, at what minimum speed must it be traveling to melt on impact? (The melting temperature of Lead is 327°C)

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Vespertilio Vespertilio · Answer 1 · 2019-11-28T07:25:02+01:00

Answer:

v = 609 m/s

Explanation:

mass of lead bullet, m = 55 g

T1 = 24°C

T2 = 327°C

Let v be the velocity

kinetic energy of bullet

K = 1/2 mv^2 = 0.5 x 0.055 x v^2

72 % of kinetic energy is used to melt the lead bullet .

Heat gained by lead = m x c x ΔT + m x L

where, c be the specific heat of lead and L be the latent heat of lead

c = 536 J/kg°C

L = 23000 J/kg

0.0275 x v^2 = 0.055 (536 x 303 + 23000)

v = 608.9 m/s

v = 609 m/s

Explanation:

okpalawalter8 okpalawalter8 · Answer 2 · 2022-04-08T10:50:23+02:00

The minimum speed must it be traveling to melt on impact is mathematically given as

v = 609 m/s

What minimum speed must it be traveling to melt on impact?

Question Parameter(s):

A 55.0-gram lead bullet hits a steel plate, both initially at 24 °C. The bullet melts and splatters on impact.

the bullet receives 72% of its kinetic energy.

Generally, the equation for the kinetic energy is mathematically given as

K = 1/2 mv^2

Therefore

K= 0.5 x 0.055 x v^2

Where Heat gained by lead

Hl= m x c x ΔT + m x L

Therefore

0.0275 x v^2 = 0.055 (536 x 303 + 23000)

v = 609 m/s

In conclusion

v = 609 m/s