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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 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.

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usmanbiu

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 [tex]v^{2}[/tex]

kinetic energy = 0.5 x 0.0046 x [tex]278^{2}[/tex]

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

okpalawalter8

The temperature change is mathematically given as

dT = 302 °c

What is its temperature change?

Question Parameter(s):

A 4.6 g lead bullet moving at 278 m/s strikes a steel plate and stops

Assume the specific heat of lead is 128 J/kg · ◦ C...

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

E = m x c x ΔT

Therefore

177.8 = 0.0046 x 128 x dT

dT = 177.8 / 0.59

dT = 302 °c

In conclusion

dT = 302 °c

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