What would be the final equilibrium temperature of 80.0g of aluminum at 5.0 c having a specific heat of 0.90 is placed in a 100.0 grams of water having a temperature of 60.0 grams?

When

Mw*Cw* ΔTw = M(Al) * C(Al) * ΔT(Al)

when we have Mw (mass of water) = 100g 

and Cw (specific heat of water) = 4.18

ΔTw (difference in temperature) = (60-Tf)

and M(Al) (mass of Al ) = 80 g

and C(Al) = 0.9

and ΔT (difference in temperature) = (Tf- 5)

as we see here, the water loses heat and the Al will gain it

so, by substitution:

∴ 100 * 4.18 * (60 -Tf) = 80 * 0.9 * ( Tf - 5 )

∴Tf = 51.9 °C  

Ais your answer to this