A225-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

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