Calculating Temperature Change in the Bullet and Block Problem

[avinhajo]

Homework Statement

A 0,47 g lead bullet moving at 151 m/s is stopped in a block of wood. All of the kinetic energy goes into heat energy added to the bullet. The initial temperature of the bullet is 32 degrees C. What is the final temperature of the bullet in degrees Celsius. The specific heat of the bullet is 130 and the mass of the wooden block is 5 g

Relevant Equations

Mv^2/2

I think that the kinetic energy is equal to heat right. Mv^2 / 2 = ms delta T
Delta T is 0,64
But I don’t really know if the temperature decreases or increases at the end

 

[osilmag]

Welcome. Do you need help?

 

[DaveC426913]

Type the words you want to say on the keyboard.

 

[avinhajo]

Type the words you want to say on the keyboard.

Now I know how this works haha

 

[avinhajo]

Welcome. Do you need help?

Yes please

 

[DaveC426913]

You need to show your attempts at an answer before we can help.

 

[avinhajo]

You need to show your attempts at an answer before we can help.

Oh ok, I will

 

[PeroK]

Homework Statement:: A 0,47 g lead bullet moving at 151 m/s is stopped in a block of wood. All of the kinetic energy goes into heat energy added to the bullet. The initial temperature of the bullet is 32 degrees C. What is the final temperature of the bullet in degrees Celsius. The specific heat of the bullet is 130 and the mass of the wooden block is 5 g
Relevant Equations:: Mv^2/2

I think that the kinetic energy is equal to heat right. Mv^2 / 2 = ms delta T
Delta T is 0,64
But I don’t really know if the temperature decreases or increases at the end

Is that the whole question? Is there a diagram that shows whether the block can move or not?

Are you sure that the mass of the block of wood is not $5kg$?

You might want to change the title from "Physics" to "Bullet and Block problem".

 

[avinhajo]

Is that the whole question? Is there a diagram that shows whether the block can move or not?

Are you sure that the mass of the block of wood is not $5kg$?

You might want to change the title from "Physics" to "Bullet and Block problem".

Yes, the mass of the wooden block is 5 g NOT 5kg.
There is no diagram but the bullet get stuck in the block so it’s an inelastic collusion I think. I’ve calculated the velocity of both bullet and block and got 12.97 m/s

 

[PeroK]

Yes, the mass of the wooden block is 5 g NOT 5kg.
There is no diagram but the bullet get stuck in the block so it’s an inelastic collusion I think. I’ve calculated the velocity of both bullet and block and got 12.97 m/s

I don't know that I would trust $5g$ of wood to stop a bullet. The $13m/s$ might be right (assuming the bullet does stop) but it is still quite fast.

 

[avinhajo]

if we assume that it’s correct, the question would be if the temperature of the bullet increases or decreases at the end? The temperature difference is 0,64. So should the temperature of the bullet be 32.64 degrees Celsius or 31.36 Celsius?

 

[osilmag]

I would think that the change in energy would flow from the speeding bullet to the block/bullet mass, bringing the temperature of the bullet down, and the block up.

 

[PeroK]

if we assume that it’s correct, the question would be if the temperature of the bullet increases or decreases at the end? The temperature difference is 0,64. So should the temperature of the bullet be 32.64 degrees Celsius or 31.36 Celsius?

If you want to cool down a bullet you put it in the fridge! You don't fire it into a block!

 

[avinhajo]

I would think that the change in energy would flow from the speeding bullet to the block/bullet mass, bringing the temperature of the bullet down, and the block up.

Even though the question says “All of the kinetic energy goes into heat energy added to the bullet”.

 

[PeroK]

I would think that the change in energy would flow from the speeding bullet to the block/bullet mass, bringing the temperature of the bullet down, and the block up.

You think wrong! Certainly the block ought to heat up as well, but the simplifying assumption is that the block does not heat up.

 

[avinhajo]

Thank

If you want to cool down a bullet you put it in the fridge! You don't fire it into a block!

Oh now I know, thank you so much

 

[PeroK]

if we assume that it’s correct, the question would be if the temperature of the bullet increases or decreases at the end? The temperature difference is 0,64. So should the temperature of the bullet be 32.64 degrees Celsius or 31.36 Celsius?

The temperature increase must be much greater than that.

 

[avinhajo]

The temperature increase must be much greater than that.

Hmm what have I done wrong?

 

[PeroK]

Hmm what have I done wrong?

It's hard to say since you haven't shown your working.

 

[avinhajo]

It's hard to say since you haven't shown your working.

 

[PeroK]

How much KE was lost by the bullet?

 

[avinhajo]

5.3 J?

 

[PeroK]

5.3 J?

That sounds about right. How does that translate into a temperature increase?

 

[avinhajo]

That sounds about right. How does that translate into a temperature increase?

I don’t think I understand

 

[PeroK]

I don’t think I understand

You must have understood something to compute a temperature increase of $0.64K$.

 

[avinhajo]

You must have understood something to compute a temperature increase of $0.64K$.

Kinetic energy = heat energy?

 

[avinhajo]

You must have understood something to compute a temperature increase of $0.64K$.

Is it 5,3 = ms delta T?

 

[haruspex]

All of the kinetic energy goes into heat energy added to the bullet.

I assume this should say "All of the lost kinetic energy goes into heat energy added to the bullet. "

 

[avinhajo]

I assume this should say "All of the lost kinetic energy goes into heat energy added to the bullet. "

Oooohhh I see. Thank you so much

 

[PeroK]

Is it 5,3J = ms delta T?

Yes, so let's see the calculation. I assume $s = 130 J kg^{-1}K^{-1}$.