Thermochemistry practice problem

Question

A 5.00-g sample of aluminum pellets (specific heat capacity "

0.89 J/'C ! g) and a 10.00-g sample of iron pellets (specific heat

capacity " 0.45 J/'C ! g) are heated to 100.0'C. The mixture

of hot iron and aluminum is then dropped into 97.3 g water at

22.0'C. Calculate the final temperature of the metal and water

mixture, assuming no heat loss to the surroundings.

Answer 1

heat lost by metal = heat gained by water

q = mC∆T

q = heat

m = mass

C = specific heat

∆T = change in temperature

For the metals:

Al: q = (5.00 g)(0.89 J/g/degree)(100 - Tf)

Fe: q = (10.00 g)(0.45 J/g/degree)(100 - Tf)

For water:

q = (97.3 g)(4.184 J/g/degree)(Tf - 22)

heat lost by Al and Fe = heat gained by water:

(5.00 g)(0.89 J/g/degree)(100 - Tf) + (10.00 g)(0.45 J/g/degree)(100 - Tf) = (97.3 g)(4.184 J/g/degree)(Tf - 22)

445 - 4.45Tf + 450 - 4.5Tf = 407Tf - 8956

415.95Tf = 9851

Tf = 23.7ºC