Where Does the Energy Go When a Ball Becomes Embedded?

AI Thread Summary
When a ball bearing drops and becomes embedded in the ground, its gravitational potential energy (GPE) is converted into kinetic energy during the fall. Upon impact, the kinetic energy is not fully converted back into GPE since the ball does not bounce, leading to energy conservation in other forms. The internal energy of both the ball bearing and the ground increases due to deformation and heat generation at the point of impact. To calculate the temperature increase of the ball bearing, one must consider the absorbed internal energy and the specific heat capacity of the material. The discussion highlights the importance of energy transformation and conservation in understanding the behavior of the ball bearing upon impact.
barnacksam
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Ive got this question but i don't know how to answer it.

A steam boiler produces steam at a rate of 0.5 kg.s-1. If the specific latent heat of vaporisation of water is 2.3 MJ kg-1:

(a) What power is consumed to achieve this rate?

cheers
 
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You have the mass flow rate and the specific energy required, so all you need to do is multiply the mass flow rate by the specific energy to get a result of MJ/s, also known as a megawatt (MW).

\frac{kg}{s} * \frac{MJ}{kg} = \frac{MJ}{s} = MW
 
thanks a lot
 
Just working through a past paper atm, just need a hand with this one as well ha

A ball bearing, of mass 200 g, is dropped from the top of a 200 m building; it falls to the ground and becomes embedded.

(a) Describe the energy changes which take place.
(5 marks)

(b) What is the increase in internal energy of the ground and the ball bearing at the point of impact?
(5 marks)

(c) Calculate the temperature increase of the ball bearing if it absorbs 70% of this internal energy.
(5 marks)

State all assumptions made.

Take the specific heat capacity of the metal to be 460 J kg-1 K-1.
 
Why don't you show your own work at an attempted solution? Pay attention to the forms of energy are being presented in the system...
 
Ive gone from GPE to kinetic, then i don't understand what happens when it becomes embedded.
then the b) and c) i just don't have a clue
 
"Becomes embedded" is important because it means the ball doesn't bounce- hence all of that energy is conserved in other areas. If the energy isn't used to bounce the ball, where does it go?
 
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