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

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To find the final temperature of a 97.6-g lead ball dropped from 4.57 m, the potential energy (P.E.) is calculated using the formula P.E. = mgh, resulting in 4.371 J. This energy is assumed to convert entirely into heat, which is expressed using the equation Q = mcΔT. The user sets up the equation 4.371 J = (0.0976 kg)(128 J/kg°C)(Tf - 20°C) to solve for the final temperature (Tf). The discussion emphasizes the importance of following forum rules and providing initial attempts for assistance. The method outlined is correct and should yield the desired final temperature.
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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
 
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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

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.
 
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
 
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
That should work.
 
Thanx!
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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