A45-g aluminum spoon (specific heat 0.88 j/g °c) at 24 °c is placed in 180 ml (180 g) of coffee at 85 °c andthe temperature of the two become equal.(a) what is the final temperature when the two become equal? assume that coffee has the same specific heat aswater.

When Heat lost by coffee -q= m( coffee) * C (coffee) * ΔT

and Heat gained by spoon q = m (spoon) * C (spoon) * ΔT

when q = -q 
∴ m(coff)* C(coff) * ΔT (coff) = m(sp)* C(sp)*ΔT(sp)
when the final tempreature of both will be the same so,
m(coff) * C(coff) * ( Tf - Ti(coff)) = m(sp) * C(sp) * (Tf- Ti(sp)

by substitution:
note: as the coffe will reduce heat so, ΔT = (85-Tf)
and the spoon will gain heat so ΔT = (Tf-24)
180 g * 4.186 * (85-Tf) = 45 * 0.88 * (Tf- 24)
∴Tf = 81 °C 

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