Answer:
The minimum speed of a lead bullet is 273.9 m/s
Explanation:
Given that,
Initial temperature = 29.5°C
We need to calculate the heat
Using formula of heat
[tex]Q= mc(T_{2}-T_{1})+mL[/tex]
Where, [tex]mc(T_{2}-T_{1})[/tex] = Energy needed to bring bullet from 29 to 327.3°
ml=Energy needed to melt bullet at this temperature
Put the value into the formula
[tex]Q=m(126\times(327.3-29.5))+m\times2.1\times10^{-4}[/tex]
We need to calculate the minimum speed of a lead bullet
Kinetic energy converted to heat energy
[tex]Q=\dfrac{1}{2}mv^2[/tex]
Put the value into the formula
[tex]m(126\times(327.3-29.5))+m\times2.1\times10^{-4}=\dfrac{1}{2}mv^2[/tex]
[tex]v^2=2(126\times(327.3-29.5)+2.1\times10^{-4})[/tex]
[tex]v=\sqrt{2(126\times(327.3-29.5)+2.1\times10^{-4})}[/tex]
[tex]v=273.9\ m/s[/tex]
Hence, The minimum speed of a lead bullet is 273.9 m/s