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

1. ### Senjai

104
[SOLVED] Q=mcT heat problem, were did i go wrong?

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
$$Q = mc \Delta T$$

3. The attempt at a solution
i found in my textbook that aluminum has a specific heat capacity of 900
so c = 900.

$$Q = mc \Delta T$$
$$Q = mc( T' - T )$$
$$T' - T = \frac{Q}{mc}$$
$$T' = \frac{Q + T}{mc}$$
$$T' = \frac{(6.0 x 10^3 J) + 22\deg}{(0.045kg)(900)}$$

2. ### Senjai

104
i got 148 Deg C which is wrong... i need 170 Deg C

3. ### Kurdt

4,941
Staff Emeritus
You've just rearranged wrongly. Its not (Q+T)/mc.

4. ### Senjai

104
so is it $$T' = \frac{Q}{mc} + T$$ ??

5. ### Kurdt

4,941
Staff Emeritus
That should work.

6. ### FOrzaTR1923

1
Re: [SOLVED] Q=mcT heat problem, were did i go wrong?

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