What is the final temperature of a lead ball dropped from a height of 4.57 m?

[Silverbolt]

A 97.6-g lead ball with an initial temperature of 20°C, is dropped from rest from a height of 4.57 m. The collision between the ball and ground is totally inelastic. Assuming all the ball's kinetic energy goes into heating the ball, find its final temperature.

PLEASE HELP ME UNDERSTAND THIS PROBLEM STEP-BY-STEP

 

[gneill]

A 97.6-g lead ball with an initial temperature of 20°C, is dropped from rest from a height of 4.57 m. The collision between the ball and ground is totally inelastic. Assuming all the ball's kinetic energy goes into heating the ball, find its final temperature.

PLEASE HELP ME UNDERSTAND THIS PROBLEM STEP-BY-STEP

You'll have to re-read the forum rules. You need to follow the posting template and provide an attempt so that we can know how to help you; we cannot provide answers or do your homework for you.

 

[Silverbolt]

This is what I think I would do:
So I would us the potential energy equation: P.E.=mgh
so --> P.E.= (.0976 kg)(9.8m/s^2)( 4.57m)
P.E. = 4.371J

With this I'll use the equation: Q=mcT

4.371J =(.0976kg)(128 J/Kg(°C) )( Tf-20°C)

Ans so I'll solve for TfSO IS THIS RIGHT?? PLEASE HELP

 

[gneill]

This is what I think I would do:
So I would us the potential energy equation: P.E.=mgh
so --> P.E.= (.0976 kg)(9.8m/s^2)( 4.57m)
P.E. = 4.371J

With this I'll use the equation: Q=mcT

4.371J =(.0976kg)(128 J/Kg(°C) )( Tf-20°C)

Ans so I'll solve for Tf


SO IS THIS RIGHT?? PLEASE HELP

That should work.

 

[Silverbolt]

Thanx!