The final temperature of a 97.6-g lead ball dropped from a height of 4.57 m can be calculated using the principles of energy conservation. The potential energy (P.E.) of the ball is determined using the formula P.E. = mgh, resulting in 4.371 J. This energy is converted into heat, which can be calculated using the equation Q = mcΔT, where m is the mass, c is the specific heat capacity of lead (128 J/kg°C), and ΔT is the change in temperature. Solving for the final temperature (Tf) yields a definitive result.
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Silverbolt said: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
That should work.Silverbolt said: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