Chapter 12, Problem 61A

To determine

The temperature change in the lead bullet.

Answer to Problem 61A

The temperature change in the lead bullet is $290.86°\text{C}$.

Explanation of Solution

Given:

A lead bullet of mass 4.2-g moving with a velocity of 275 m/s strikes a steel plate and the kinetic energy of the bullet is converted into thermal energy without any loss and hence increases the temperature of the bullet.

Formula Used:

According to First Law of thermodynamics, the thermal energy change of an object is,

  $\Delta U=Q-W$

Where,

Q is the heat supplied to the object

W is the work done by the object.

Now, when the system is isolated from the surroundings, the thermal energy is

  $\Delta U=-W$........(1)

Here, the work done by the bullet is converted to the thermal energy.

The work done by the bullet of mass m is its change in kinetic energy, expressed as,

  $\begin{array}{c}{W}_{bullet}=\Delta K.E\\ =\frac{1}{2}m{\left(\Delta v\right)}^2\end{array}$

In the above equation, the parameter $\Delta v$ represents the change in velocity.

The energy for causing temperature change is dependent on the temperature change, mass of the object $m$ and specific heat $C$ , and is given by the expression,

  $Q_t=mC\Delta T$

Calculation:

Here, bullet is having a mass of $m\text{=4}\text{.2 - g}$ .The specific heat of lead is $C_l=130\text{ J/kg}\text{.K}$ . It moves with a velocity $v\text{=275 m/s}$ and comes to rest making its final velocity to be zero. Now, the change in kinetic energy of the bullet is given by,

  $\begin{array}{c}\Delta K.E=\frac{1}{2}m{\left(\Delta v\right)}^2\\ =\frac{1}{2}\times 4.2\times {10}^{-3}\times {275}^2\\ =158.8125\text{ J}\end{array}$

The thermal energy for the bullet can be calculated as,

  $\begin{array}{c}\Delta U=m{C}_l\Delta T\\ =4.2\times {10}^{-3}\times 130\times \Delta T\\ =0.546\Delta T\end{array}$

Now, the change in Kinetic energy is converted into thermal energy and hence it follows that,

  $\begin{array}{c}\Delta U=\Delta K.E\\ 0.546\Delta T=158.8125\\ \Delta T=\frac{158.8125}{0.546}\\ =290.86°\text{C}\end{array}$

Conclusion:

Therefore, the temperature change in the bullet is $290.86°\text{C}$ .