A 225-g sample of aluminum was heated to 125.5 oc, then placed into 500.0 g water at 22.5 oc. (the specific heat of aluminum is 0.900 j/goc). calculate the final temperature of the mixture. (assume no heat loss to the surroundings.)

when heat gained = heat lost

when AL is lost heat and water gain heat

∴ (M*C*ΔT) AL = (M*C*ΔT) water

when M (Al) is the mass of Al = 225g

C (Al) is the specific heat of Al = 0.9

ΔT (Al) = (125.5 - Tf)

and Mw is mass of water = 500g

Cw is the specific heat of water = 4.81

ΔT = (Tf - 22.5)

so by substitution:

∴225 * 0.9 * (125.5 - Tf) = 500 * 4.81 * (Tf-22.5)

∴Tf = 30.5 °C