Question

A 17.0 g bullet traveling horizontally at 785 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? (in degrees)​

Answer 1

$\red{\bold{{\underline{ Answer \: with\: Explanation \: :}}}}$

The Maximum Temperature increase is $\Delta T = 0.0497 \ ^oC$

From the question we are told that

    The mass of the bullet is $m = 17.0 \ g =0.017 \ kg$

     The  speed is  $v_1 = 785 \ m/s$

     The mass of the water is  $m_w = 13.5 \ kg$

     The velocity it emerged with is  $v_2 = 534 \ m/s$

Generally due to the fact that energy can nether be created nor destroyed but transferred from one form to another then  

The Change in kinetic energy of the bullet =  The Heat gained by the water

 So

 The change in kinetic energy of the water is  

          $\Delta KE = \frac{1}{2} m (v_1^2 - v_2 ^2 )$

substituting values  

        $\Delta KE =0.5 * 0.017 * (( 785)^2 - (534) ^2 )$

        $\Delta KE = 2814.1 \ J$

Now the heat gained by the water is

     $Q = m_w* c_w * \Delta T$

Here $c_w$ is the specific heat of water which has a value  $c_w = 4190 J/kg \cdot K$

So  since   $\Delta KE = Q$  

we have that

          $2814.1 = 13.5 * 4190 * \Delta T$

          $\Delta T = 0.0497 \ ^oC$