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

by Senjai
Tags: heat, qmct, solved
 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 $$Q = mc \Delta T$$ 3. The attempt at a solution i found in my textbook that aluminum has a specific heat capacity of 900 so c = 900. $$Q = mc \Delta T$$ $$Q = mc( T' - T )$$ $$T' - T = \frac{Q}{mc}$$ $$T' = \frac{Q + T}{mc}$$ $$T' = \frac{(6.0 x 10^3 J) + 22\deg}{(0.045kg)(900)}$$
 P: 104 Q=mcT heat problem, were did i go wrong? so is it $$T' = \frac{Q}{mc} + T$$ ??