Finding temperature after something falls

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crazyog
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Homework Statement


What is the temperature in K of 1.0 kg of lead initially at 300 K after it falls 200 m? The specific heat of lead is 128 J/kg C.
a) 307 b) 311 c) 315 d)275 e) 279
the answer my teacher said was c but i keep getting a.


Homework Equations


I thought I would use mc (Tf-Ti) = mgy

The Attempt at a Solution


The masses cross out. so
(300)(delta T) = (9.8)(200)
I get delta T = 6.53
300 +6.53 = 306.53
 
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(1kg)(128) (Tf - 300) = (1 kg) (9.8) (200 m)
128Tf - 38400 = 1960
Tf = 315.31
thank you! :)ehh, how do I mark this solved?
 
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