A lead bullet penetrates into a solid object and melts. Assuming that 50% of its K.E was used to heat it. Calculate the initial speed of the bullet. Initial temperature of the bullet is 270C and its melting point is3270C. Latent heat of fusion of lead is 2.5x104 J kg-1 . And sp. Heat capacity of lead is 125 J Kg-1
Answers
Given :
Initial temperature of the bullet =27°C
melting point = 327°C
latent heat of fusion of lead =2.5×10⁴ J kg−1
specific heat of lead =125J kg−1
Heat energy needed to melt bullet is Q= Q1+Q2
where
Q1= msΔT
Q1=m×125×(327−27) =3.75×10⁴m
Q2=mL=m×2.5×10⁴ =2.5×10⁴m
Q=6.25 x 10⁴ m
If v is the speed of bullet then 50% of 1/2mv² should be equal to Q.
(50/100) x 1/2 mv² =6.25x 10⁴m
v=√(6.25×10⁴×2)√/0.5 =500m/s=1800km/h
∴The initial speed of the bullet is 1800km/h
Answer:
Kinetic energy of bullet = 1/2mV2
50% of kinetic energy = 1/4MV2
50% og kinetic energy = heat required to raised its temperature by 600 K and to melt
Explaination:
Formula is -
1/4MV2=Ms dT+ML
V2/4=sdTQ+ML
V2/4=125×300+2.5×10*4
V2=10×10*4+ 15×10*5
V2= 25×10*4
V = 5×10*2 m/s
V= 1800kmphs
The initial speed of the bullet is 1800kmphs
Hope it helps you