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[SOLVED] Q=mcT heat problem, were did i go wrong?
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?
Q = mc \Delta T
i found in my textbook that aluminum has a specific heat capacity of 900
so c = 900.
Q = mc \Delta T
Q = mc( T' - T )
T' - T = \frac{Q}{mc}
T' = \frac{Q + T}{mc}
T' = \frac{(6.0 x 10^3 J) + 22\deg}{(0.045kg)(900)}
Homework Statement
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?
Homework Equations
Q = mc \Delta T
The Attempt at a Solution
i found in my textbook that aluminum has a specific heat capacity of 900
so c = 900.
Q = mc \Delta T
Q = mc( T' - T )
T' - T = \frac{Q}{mc}
T' = \frac{Q + T}{mc}
T' = \frac{(6.0 x 10^3 J) + 22\deg}{(0.045kg)(900)}