# Heat problem

1. May 29, 2006

### epsbanga12

HI!

This problem has been killing me...

The Q is the following: A canon fires a canon ball which weighs 50 grams vertically at an initial speed of 600 m/s. 3 Km on top of the point where it was fired, the speed is only 50 m/s. calculate the heat during the canon ball rise.

I dont know how to use the kinetic formula in function with the energy formula, any help would be appreciated!

2. May 29, 2006

### Mattara

The canon ball has a certain amount of kinetic energy from the start and a certain kinetic energy when the speed is 50 m/s (which is less). Some of the kinetic energy has been transformed to potential energy. The rest of the energy has been "lost" as heat etc.

Start: $$W_k$$
Finish $$W_k + W_p + W_h$$

Lets call the heat energy $$W_h$$ in this case although it isn't exactly correct index.

3. May 29, 2006

### epsbanga12

Thanks for clearing things up, Mattara.

So what would we eventually have to do to find Q, the heat produced?

4. May 29, 2006

### Hootenanny

Staff Emeritus
Using Mattara's notation Q = Wh.

~H

5. May 29, 2006

### epsbanga12

so it would be Wh + Wp = -Wh ?

6. May 29, 2006

### Hootenanny

Staff Emeritus
Not quite, intially you have some kinetic energy. At the 'end' you have some kinetic energy, some potential and the rest as heat, therefore;

Initial Kinetic = Final Kinetic + Potential Energy + Heat

$$\frac{1}{2}mv_{i}^{2} = \frac{1}{2}mv_{f}^{2} + mgh + Q$$

Can you go from here?

~H

Last edited: May 29, 2006
7. May 29, 2006

### epsbanga12

hehe thanks Hootenanny
Actually, i meant Wk + Wp= -Wh, the former Wh was a typo...

But everything is clear now, thanks a lot to both of you!