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Q=mcT heat problem, were did i go wrong?

by Senjai
Tags: heat, qmct, solved
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Senjai
#1
Dec4-07, 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]
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Senjai
#2
Dec4-07, 11:16 AM
P: 104
i got 148 Deg C which is wrong... i need 170 Deg C
Kurdt
#3
Dec4-07, 11:36 AM
Emeritus
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PF Gold
P: 4,980
You've just rearranged wrongly. Its not (Q+T)/mc.

Senjai
#4
Dec4-07, 11:38 AM
P: 104
Q=mcT heat problem, were did i go wrong?

so is it [tex] T' = \frac{Q}{mc} + T [/tex] ??
Kurdt
#5
Dec4-07, 11:38 AM
Emeritus
Sci Advisor
PF Gold
P: 4,980
That should work.
FOrzaTR1923
#6
Apr8-09, 05:28 AM
P: 1
kurdt ,
what would the worked out solution look like for the last step?


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