A 15.0 g bullet traveling horizontally at 865 m/s passes through a tank containing 13.5 kg of water and emerges with a speed of 534 m/s . What is the maximum temperature increase that the water could have as a result of this event?

The term speed may be defined as the ratio of distance and time.

Consider the given data as below.

Mass of the bullet, $\mathrm{M}=15.0\mathrm{g}=15.0\times {10}^{-3}\mathrm{kg}$

Initial velocity, ${\mathrm{v}}_1=865\mathrm{m}/\mathrm{s}$

Final velocity, ${\mathrm{v}}_2=534\mathrm{m}/\mathrm{s}$

Heat capacity of water, ${\mathrm{C}}_{\mathrm{w}}=4.19\times {10}^3\mathrm{J}/°\mathrm{C}.\mathrm{kg}$

Mass of a tank, $\mathrm{m}=13.5\mathrm{kg}$

Using the concept of kinetic energy,

$$\begin{gathered} ∆\mathrm{K}={\mathrm{K}}_{\mathrm{before}}-{\mathrm{K}}_{\mathrm{after}} \\ =\frac{1}{2}\times \mathrm{M}\times \left({\mathrm{v}}_1^2-{\mathrm{v}}_2^2\right) \end{gathered}$$

Here, $∆\mathrm{K}$ is the change in kinetic energy, ${\mathrm{K}}_{\mathrm{before}}$is the initial kinetic energy, ${\mathrm{K}}_{\mathrm{after}}$is the final kinetic energy, m is the mass, ${\mathrm{v}}_1$is the initial velocity, and ${\mathrm{v}}_2$is the final velocity.

Substitute known values in the above equation.

$$\begin{gathered} ∆\mathrm{K}=\frac{1}{2}\times 15\times {10}^{-3}\left({\left(865\right)}^2-{\left(534\right)}^2\right) \\ =3,47\times {10}^3\mathrm{J} \end{gathered}$$

That is the loss of kinetic energy which transfer to the water as a heat so the maximum temperature increase of water is,

$$\begin{gathered} \mathrm{\delta T}=\frac{\mathrm{Q}}{\mathrm{m}\left({\mathrm{C}}_{\mathrm{w}}\right)} \\ =\frac{3.47\times {10}^3\mathrm{J}}{13.5\mathrm{kg}\times 4.19\times {10}^3\mathrm{J}/°\mathrm{C}.\mathrm{kg}} \\ =0.0613°\mathrm{C} \end{gathered}$$

Hence, the maximum temperature increase is$0.0613°\mathrm{C}$.