Question

8.
A lead bullet (specific heat = 0.032 cal/gm °C)
is completely stopped when it strikes a target
with a velocity of 300 m/s. The heat generated
is equally shared by the bullet and the target
The rise in temperature of bullet will be
1) 16.7°C
2) 1.67°C
3) 167.4°C
4) 267.4°C 2​

Answer 1

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

3. 167.41

Answer 2

Answer:

167.41°C

Step-by-step explanation:

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