How much heat energy is produced when a steel ball falls into sand?

[biamin]

A steel ball falls from a height of 20m into a pile of sand. If 1/2 the energy ends up i as heat in the ball,m how much is the ball heated?

I am starting out with the work equation.
Work = force * displacement.
Work = 9.81m/s^2(20m)
Work = 196.2 m^2/s^2


I am having trouble relating this to a Joule or calorie or some type of heat energy.

Thanks for any help

 

[cepheid]

Welcome to PF biamin!

You forgot the mass (m) in F = mg. The force you used just had g in it, but no mass. So it wasn't really a force at all, and that's why the units weren't working out for you.

 

[biamin]

unfortunately the problem gives no mass for the steel ball. I am not sure how to approach the problem.

The only other formula I can think of to use is the one for potential energy (mass*gravity*height) but that has a mass in it as well.

Any thoughts?

 

[cepheid]

unfortunately the problem gives no mass for the steel ball. I am not sure how to approach the problem.

The only other formula I can think of to use is the one for potential energy (mass*gravity*height) but that has a mass in it as well.

Yeah, in fact, that's not even really a different method. After all, using work = force*distance, you'd get force = mg, and distance = h, hence work = mgh. This makes sense, since the work done by gravity is always the negative of the change in potential energy.

Any thoughts?

It doesn't seem to me like you have enough information to get a numerical answer. Express the amount of energy in terms of the unknown mass, m, and move on?

EDIT: I just had a thought. Maybe the question is asking for the temperature change of the ball?? Since the temperature change resulting from a given amount of heat also depends on m, the m's will cancel from both sides of the equation (in other words, the result will be independent of mass, since more massive balls would require more heat to increase in temperature by the same amount, but they would also have more work done on them by gravity, and these two effects would exactly cancel). So you'd have something like:

heat energy gained = 1/2 work done on ball

Q = ½W

mcΔT = ½mgh

where c is the specific heat capacity of steel in J/(g°C) , which you just have to look up. You can see that mass doesn't matter for the end result.

 

[biamin]

Thanks Cepheid,

The specific heat capacity was the bit of information I was needing to add to the equation.