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]

Mass of the bullet, $m = 4.2 \text{ g} = 4.2 \times 10^{-3} \text{ kg}$

Velocity of the bullet, $v = 275 \text{ ms}^{-1}$

Specific heat of lead, $s = 130 \text{ J/kg}^\circ \text{C}$

Kinetic energy is converted to thermal energy: $K.E. = Q$

Kinetic Energy: $K.E. = \frac{1}{2}mv^2$

Thermal Energy (Heat): $Q = ms\Delta T$

Setting the two equations equal to each other: $\frac{1}{2}mv^2 = ms\Delta T$

Notice that the mass ($m$) cancels out from both sides: $\frac{1}{2}v^2 = s\Delta T$

Now, solve for the temperature change ($\Delta T$): $\Delta T = \frac{v^2}{2s}$

Substitute the given values: $\Delta T = \frac{(275)^2}{2 \times 130}$

$\Delta T = \frac{75625}{260}$

$\Delta T \approx 290.865^\circ \text{C}$

$\Delta T \approx 291^\circ \text{C}$