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…

Question

asked Oct 13, 2021 131k views

1 Answer

Andrey Borisov · Answer 1 · 2021-10-18T09:23:34+0000

Answer:

The maximum temperature increase is
$\Delta T = 0.0497 \ ^oC$

Step-by-step explanation:

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

$\Delta KE = (1)/(2) m (v_1^2 - v_2 ^2 )$

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

$2814.1 = 13.5 * 4190 * \Delta T$

$\Delta T = 0.0497 \ ^oC$