Question

A lead bullet moving at 70 ms^-1 is brought to rest on hitting a target. If 80% of its energy is converted into heat energy, find the rise in temperature (Sp. Heat cap. Of lead is 140 J/kgK).

Answer 1

Given,initial velocity, u = 70 m/s

final velocity, v = 0 m/s

sp. heat capacity of lead,s = 140 J/kgK

Let Δt be the change in temperature

K.E possessed by the bullet = ${\displaystyle \frac{1}{2}}$ m (70)² = 2450m joules

Heat energy = 80% of K.E = ${\displaystyle \frac{80}{100}\times 2450}$m = 1960m joules

We know that, H = m s Δt

or, 1960m = m x 140 x Δt

or, Δt = ${\displaystyle \frac{1960}{40}}$ = 14 °C