Answer:
The maximum temperature increase is [tex]\Delta T = 0.0497 \ ^oC[/tex]
Explanation:
From the question we are told that
The mass of the bullet is [tex]m = 17.0 \ g =0.017 \ kg[/tex]
The speed is [tex]v_1 = 785 \ m/s[/tex]
The mass of the water is [tex]m_w = 13.5 \ kg[/tex]
The velocity it emerged with is [tex]v_2 = 534 \ m/s[/tex]
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
[tex]\Delta KE = \frac{1}{2} m (v_1^2 - v_2 ^2 )[/tex]
substituting values
[tex]\Delta KE =0.5 * 0.017 * (( 785)^2 - (534) ^2 )[/tex]
[tex]\Delta KE = 2814.1 \ J[/tex]
Now the heat gained by the water is
[tex]Q = m_w* c_w * \Delta T[/tex]
Here [tex]c_w[/tex] is the specific heat of water which has a value [tex]c_w = 4190 J/kg \cdot K[/tex]
So since [tex]\Delta KE = Q[/tex]
we have that
[tex]2814.1 = 13.5 * 4190 * \Delta T[/tex]
[tex]\Delta T = 0.0497 \ ^oC[/tex]
