18.2 grams of Aluminum and 18.2 grams of Beryllium

Question

18.2 grams of Aluminum and 18.2 grams of Beryllium both start at room temperature (25°C). If 49.4 J of heat are added to each separately, what is the temperature difference between the two after they have been heated? The specific heat for Al = 0.899 J/g°C and Be = 1.82 J/g°C.

Answer 1

Al: q = mC∆T

∆T = q/mC = 49.4 J/(18.2 g)(0.899 J/g/deg) = 3.02 degrees

Final temperature of aluminum = 25º + 3.02º = 28ºC

Be: q = mC∆T

∆T = q/mC = 49.4 J/(18.2 g)(1.82 J/g/deg) = 1.49 degrees

Final temperature of beryllium = 25º + 1.5º = 26.5º