Question

A lead bullet just melts when stopped by an obstacle. Assuming that 25 per cent of the heat is absorbed by the obstacle, find the velocity of the bullet if its initial temperature is $27^{\circ}C$. [For lead, melting point = $327^{\circ}C$, specific heat = $0.03 cal/g-^{\circ}C$, latent heat of fusion = 6 cal/g and J = 4.2 J/cal.]

Answer 1

The velocity of the bullet is V = 409.9 m/s

Explanation:

• Kinetic energy of bullet = 1/2 mv^2

Heat (Q) = 3/4 (1/2mv^2)

3/4 (1/2mv^2) = msΔT + mL

Initial temperature Ti = 27°C

Final temperature Tf = 327°C

• Now the velocity of bullet if given by.

V = √ 8/3 (SΔT + L)

• Calculate S &L

S = 0.03 x 4.2 J / KgK

L = 6 Cal / gm = 6000 x 4.2 J / K

V = √ 8/9 ( 3 x 4.2 x 300 x + 6000 x 4.2

V = 409.9 m/s

Hence the velocity of the bullet is V = 409.9 m/s

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