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April 16, 2014

April 16, 2014

Posted by **James** on Thursday, July 19, 2012 at 11:36pm.

I know the answer is 334 degree C. How do I get this answer?

- Physics -
**James**, Thursday, July 19, 2012 at 11:38pmI 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.

- Physics -
**James**, Thursday, July 19, 2012 at 11:41pmThe table gives the specific heat of lead as 0.129 KJ/(kgK) and 0.0308 cal/(gK).

- Physics -
**Elena**, Friday, July 20, 2012 at 2:33am0.7•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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