a lead bullet having an initial temperature of 30 C is fired with an initial velocity of 450 m/s. It penetrates a wall and melts. Assume that only 40% of the kinetic energy goes in heating the bullet and the rest dissipates in the wall. Latent heat of fusion of lead is 25000 J/kg and specific heat of lead is 125 J/kg/K. What is the melting point of lead?

Open in App

Solution

Dear Student,

$The heat gained by the bullet, which raises its temperature from 30 ° C to T ° C is, Q_{1} = mc ∆ t where, m = mass of bullet c = specific heat of lead = 125 J / kg / K T = melting point of lead ∆ T = rise in temperature of the bullet = T - 30 \Rightarrow Q_{1} = m \times 125 \times (T - 30) Cal . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (1) The heat required for melting the bullet is, Q_{2} = mL where, L = latent heat of fusion of lead = 2.5 \times 10^{4} J / kg \Rightarrow Q_{2} = m \times 2.5 \times 10^{4} Cal . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (2) Kinetic energy of the bullet is given by, KE = \frac{1}{2} {mu}^{2} where, u = initial velocity of bullet = 450 m / s The total amount of developed heat is, Q = 40 % \times KE Q = \frac{40}{100} \times \frac{1}{2} \times m \times {(450)}^{2} Cal . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (3) At thermal equilibrium, we can write, Q = Q_{1} + Q_{2} \frac{40}{100} \times \frac{1}{2} \times m \times {(450)}^{2} = m \times 125 \times (T - 30) + m \times 2.5 \times 10^{4} 45 \times 45 \times 20 = 125 \times (T - 30) + 25000 125 \times (T - 30) = 40500 - 25000 (T - 30) = \frac{15500}{125} T = 124 + 30 = 154 ° C$

Suggest Corrections

Similar questions

Q. A lead bullet penetrates into a solid object and melts. Assuming that

50

% of its kinetic energy was used to heat it, the initial speed of the bullet is (the initial temperature of the bullet is

25^{o} C

and its melting point is

300^{o} C

). Latent heat of fusion of lead

= 2.5 \times 10^{4} J / k g

and specific heat capacity of lead

= 125 J / k g - K

A lead bullet penetrates into a solid object and melts. Assuming that 50% of its kinetic energy was used to heat it, calculate the initial speed of the bullet. The initial temperature of the bullet is $27^{\circ} C$ and its melting point is $327^{\circ} C$ . Latent heat of fusion of lead = $2.5 \times 10^{4} J k g^{- 1}$ and specific heat capacity of lead = $125 J k g^{1} K^{1}$ .