Chemistry - Calorimetry?

Question

A 5.00-g sample of aluminum pellets (specific heat capacity = 0.89 J/C x g) and a 10.00-g sample of iron pellets (specific heart capacity = 0.45 J/C x g) are heated to 100.0C. The mixture of hot iron and aluminum is then dropped into 97.3g of water at 22.0C. Calculate the final temperature of the meta and water mixture, assuming no heat lost to the surroundings.

Answer 1

*...Heat lost by Al: (T = Final temp.)

= 5g x 0.89J/g/°C x (100 - T) = (445 - 4.45T).

*...Heat lost by Fe.

= 10g x 0.45J/g/°C x (100 - T) = (450 - 4.5T).

*...Heat Gained by water.

= 97.3g x 4.184J/g/°C x (T - 22) = (407.1T - 8956.3).

*...Total Heat lost = Total Heat gained.

(895 - 8.95T) = (407.1T - 8,956.3)

(895 + 8,956.3) = (8.95T + 407.1T)

*...T = 9,851.3 / 416.05 = 23.7°C Final temp. (Say 24°C).