Question

A lead bullet of specific heat 0.032 kcal kg-1C-1strikes a target with velocity of 300 ms-1.If the bullet is completely stopped by target ,find the rise in temperature of the bullet.Assume that the heat produced is equally shared by the target and the bullet.(J=4.2 J cal​-1). ANS:167.4 C

Answer 1

Dear Student,

Let m be the mass of the bullet.The kinetic energy associated with the bullet,K.E=12mv2K.E=12×m×(300)2=m×45000 JAccording to law of conservation of energy,m×450002=m×4.2×103×0.032×θwhere, θ=rise in temperatureθ=450002×4.2×103×0.032
=167.41°C

Answer 2

Answer:

167.4

Explanation:

formulae

1/2mvsquare =j(msdelta theta)