Take the → Table of specific thermal capacities by weight to determine the temperature attained by one mole of aluminium heated by one kilocalorie.
$Q$ $=$ $c_{Al}\cdot m\cdot (\theta_f$ $ -$ $ \theta_i)$ Let be $x$ the final temperature: $1\cdot 4184$ $=$ $890\cdot 0.02698\cdot(x$ $-$ $20)$ $x$ $\approx$ $194^oC$