∙A 15.0 g bullet traveling horizontally at 865 m / s passes through a tank containing 13.5 kg of

step 1 Okay, so the question was asking for the changing temperature. Well, we know the initial speed w

∙A 15.0 g bullet traveling horizontally at 865 m / s passes...

\bullet A 15.0 g bullet traveling horizontally at 865 $\mathrm{m} / \mathrm{s}$ passes through a tank containing 13.5 $\mathrm{kg}$ of water and emerges with a speed of 534 $\mathrm{m} / \mathrm{s}$ . What is the maximum temperature increase that the water could have as a result of this event?

Key Concepts

Specific Heat Capacity

Specific heat capacity is a property of a substance that indicates how much energy is required to raise the temperature of a unit mass of a substance by one degree Celsius. Water's specific heat capacity allows us to determine how a given amount of heat energy will affect its temperature change.

Energy Transfer in Collisions

This concept involves the process in which energy is transferred from one object to another during interactions such as collisions or passages through mediums. In problems like this, it is assumed that the kinetic energy lost by the moving object is transferred entirely as heat to the other substance, leading to a measurable temperature increase.

Conservation of Energy

This principle states that energy cannot be created or destroyed, only transformed from one form to another. In the context of the problem, the loss of kinetic energy from the bullet as it passes through water is assumed to be converted entirely into thermal energy, thereby increasing the temperature of the water.

Kinetic Energy

Kinetic energy is the energy possessed by an object due to its motion, calculated using the formula KE = 0.5 * m * v^2. The change in the bullet's kinetic energy before and after passing through the water provides the amount of energy available to be transferred into the water as heat.

Transcript

00:01 Okay, so the question was asking for the changing temperature.

00:03 Well, we know the initial speed was 865 meter per second, and the final speed is 534 meter per second.

00:10 The mass for the bullet was 15 grams, and the mass for the water is 13 .5 kilograms.

00:19 Also, we know another thing, which is the specific heat of water.

00:23 It's 4190 -0 over kilogram times kelvin.

00:30 Okay? so, well, let's take a look at the question and let me see how to do this.

00:45 Well, in order to figure out data t, we got to know q as well.

00:50 So q is unknown here.

00:52 But we can figure out q through the equation for kinetic energy, which is ke.

01:04 And how do we figure out the kinetic energy? the kinetic energy, remember it was from the bullet, the bullet was shooting from the gun and reached the water with the final speed of 534 meter per cent.

01:20 So that means the kinetic energy transfer into the heat during the process.

01:27 And due to the energy conservation law, there's no energy loss.

01:31 So basically all the kinetic energy that was burst from the bullets is converted into the heat.

01:39 In water, okay? so that's our basic idea behind this question.

01:46 Okay? so let's try to figure out the kinetic energy first, which is one -half mass of the bullet times the final speed square minus the initial speed square.

02:03 So that's the change in kinetic energy.

02:07 And let's go to the next page and they'll give us k .e.

02:12 Is equal to one -half, one -half, and then we'll give us k.

02:12 E.

02:12 Is equal to one -half, was 15 grams and 15 grand and 15 grand is equal to 15 times 10 to power 93 kilograms.

02:23 And times final speed was a6, i'm sorry.

02:31 It was 534 meter square, meter second square, okay? minus a65 meter per second square...