Question

6. A bullet of mass 20 g strikes a target with a
velocity 100 m/s and comes to rest. 50% kinetic
energy of the bullet raises its temperature by
25°C. Calculate: (i) increase in the internal
energy of the bulle temperature,

Ans. (i) 50 J, ​

Answer 1

Answer:

According to the statement of the question:

50% of K.E = msθ

1/2 [ 1/2 mV^2] = msθ

Now s = V^2/4θ

        s = (100)^2/ 4 x 298

       s = 10,000 / 1192 = 8. 389 JKg^-1 K^-1

From first law of thermodynamics:

ΔU = Δq + w

w = pΔv   [ Δv = 0]

Thus

ΔU = Δq = msθ

     = 20/100 x 8. 389 x 298

ΔU = 499.9 J

Increase in the value of internal energy is 499.9 J