Posted by James on Thursday, July 19, 2012 at 11:36pm.
I tried using:
.7E = 1/2mv^2 - cm(dT)
However, I do not know the total energy of the system, so I can't finish the problem this way.
The table gives the specific heat of lead as 0.129 KJ/(kgK) and 0.0308 cal/(gK).
0.7•KE=Q
0.7•mv²/2=mc(tº-20º)
tº= 20º+0.7•v²/2 =
=20º+0.7•340²/2•129= 333.6ºC≈334ºC
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