Question

A 15 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?

Answer 1

Answer:

0.0613°C

Explanation:

the given parameters are m=15gm=15×10⁻³  V₁=865m/s  V₂=534m/s

the bullet moves with different kinetic energies before and after the penetration, therefore

Kinetic energy before - kinetic energy after = 1/2 × m × ( V₁² - V₂²)

                                                                         =$\frac{1}{2}$ × 15×10⁻³ × (865² - 534²)

                                                                         = 3.47 × 10⁻³J

 this loss in energy is transferred to the water, therefore

change in temperature = $\frac{Q}{m C}$

where c = heat capacity of water = 4.19 x 10^3

          m = mass of water = 13.5 kg

= {3.47 × 10⁻³} / {13.5 x  4.19 x 10^3 }

=0.0613°C