Hello again,

I'm trying to figure out this problem-

A 2.4 kg iron ball is dropped from a height of 14 m onto a concrete roadway, and 2.5% of its kinetic energy at the time it reaches the ground is transformed into internal energy in the ball itself. (The rest of the energy is transmitted to the ground, converted into sound energy, and so on.) What is the ball's increase in temperature?

So far I've used the equations PE=mgh and Q=cm△T, with c=449, m=2.4kg, h=14m, g=9.8.

My work: PE=mgh=24(9.8)(14)=3292.8 J.

Q=PE/0.025

PE=0.025cm△T.=

3292.8=0.025(449)(2.4)△T.=

3292.8=26.94△T

△T=12.2K

The answer key says its 7.6e-3 K, so I tried again with 25%(329.28)= 8.23 and tried 329.28=8.23(449)(2.4)△T. Needless to say it's still wrong, and I don't know what to do with it now

Any help would be appreciated!