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
