A 4.2g lead bullet moving at 275ms^-1 strikes a steel plate and stops. If all its kinetic energy is coverted to thermal energy and none leaves the bullet, what is its temperature change? [Specific heat of lead is 130 j/kg^oC]

The kinetic energy of the bullet is given by:

$K E=\frac{1}{2} m v^{2}$

Where:

- $m=4.2 g=0.0042 \mathrm{~kg}$ (mass of the bullet)

- $v=275 \mathrm{~m} / \mathrm{s}$ (initial velocity of the bullet)

$\begin{array}{c} K E=\frac{1}{2} \times 0.0042 \times(275)^{2} \\ K E=0.0021 \times 75625 \\ K E=158.8 \mathrm{~J} \end{array}$

The heat energy gained by the bullet is:

$Q=m c \Delta T$

Where:

- $Q=158.8 \mathrm{~J}$ (from KE calculation)

- $\quad c=130 \mathrm{~J} / \mathrm{kg}^{\circ} \mathrm{C}$ (specific heat capacity of lead)

- $m=0.0042 \mathrm{~kg}$

- $\Delta T=$ temperature change

Rearranging for $\Delta T$ :

$\begin{array}{c} \Delta T=\frac{Q}{m c} \\ \Delta T=\frac{158.8}{0.0042 \times 130} \\ \Delta T=\frac{158.8}{0.546} \\ \Delta T \approx 291^{\circ} \mathrm{C} \end{array}$