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Q=mcT heat problem, were did i go wrong? 
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#1
Dec407, 11:15 AM

P: 104

1. The problem statement, all variables and given/known data
If a 45g sample of aluminum at 22 degrees C is given 6.0 x 10^3 J of heat, what will its final temperature be? 2. Relevant equations [tex]Q = mc \Delta T[/tex] 3. The attempt at a solution i found in my textbook that aluminum has a specific heat capacity of 900 so c = 900. [tex]Q = mc \Delta T [/tex] [tex]Q = mc( T'  T ) [/tex] [tex]T'  T = \frac{Q}{mc} [/tex] [tex]T' = \frac{Q + T}{mc}[/tex] [tex]T' = \frac{(6.0 x 10^3 J) + 22\deg}{(0.045kg)(900)} [/tex] 


#2
Dec407, 11:16 AM

P: 104

i got 148 Deg C which is wrong... i need 170 Deg C



#3
Dec407, 11:36 AM

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PF Gold
P: 4,980

You've just rearranged wrongly. Its not (Q+T)/mc.



#4
Dec407, 11:38 AM

P: 104

Q=mcT heat problem, were did i go wrong?
so is it [tex] T' = \frac{Q}{mc} + T [/tex] ??



#5
Dec407, 11:38 AM

Emeritus
Sci Advisor
PF Gold
P: 4,980

That should work.



#6
Apr809, 05:28 AM

P: 1

kurdt ,
what would the worked out solution look like for the last step? 


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