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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?

Respuesta :

adetoseen

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)

                                                                         =[tex]\frac{1}{2}[/tex] × 15×10⁻³ × (865² - 534²)

                                                                         = 3.47 × 10⁻³J

 this loss in energy is transferred to the water, therefore

change in temperature = [tex]\frac{Q}{m C}[/tex]

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

Cricetus

The maximum temperature increase will be "0.062°C".

Given values:

  • Mass, m = 15 g
  • Emerging speed, v = 534 m/s
  • Horizontal speed, u = 865 m/s

According to the question,

The loss in KE of bullet will be:

= [tex]Increase \ in \ heat \ content \ of \ water+Loss \ in \ conversion[/tex]

then,

→ [tex]0.5(u^2-v^2)= mw\times cp\times \Delta T+loss[/tex]

→ [tex]mw\times cw\times \Delta T=0.5\times (u^2-v^2)-loss[/tex]

hence,

The maximum increase in temperature will be:

→ [tex]mw\times cw\times (\Delta T)max = 0.5(u^2-v^2)[/tex]

→                    [tex](\Delta T)max = 0.5\times[ \frac{(u^2-v^2)}{mw\times cw} ][/tex]

By substituting the values, we get

                                      [tex]= 0.5\times 0.015[\frac{(865^2-534^2)}{13.5\times 4180} ][/tex]

                                      [tex]= 0.062^{\circ} C[/tex]

Thus the above answer is right.

Learn more about temperature here:

https://brainly.com/question/24279333

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