A 5.00 g sample of aluminum pellets and a 10.00 g sample of ...

Question

A 5.00 g sample of aluminum pellets and a 10.00 g sample of iron pellets are placed together in a dry test tube, and the test tube is heated in a boling water bath to 100°C. The mixture of hot iron and aluminum is then poured into 97.3 g of water at 20.5°C. To what final temperature is the water heated by the metals?

Answer 1

Specific heat for aluminum is 0.902 J/g/deg

Specific heat for iron is 0.450 J/g/deg

heat lost by metal = heat gained by water

(5.00 g)(0.902 J/g/deg)(∆T_metal) + (10.00 g)(0.450 J/g/deg)(∆T_metal) = (97.3g)(4.184 J/g/deg)(∆T_water)

∆Tmetal (4.51J/deg + 4.50 J/deg) = 407 J/deg ∆T

100-Tf (9.01) = 407 (Tf-20.5)

100 - 9.01Tf = 407Tf - 8344

8444 = 41 6.01Tf

Tf = 20.3 degrees C

(Do please check my math)