Answer
(a) $0.22C^{\circ}$
(b) less than
Work Step by Step
(a) We know that:
heat gained by ice cube=heat lost by Al cup and water
$Q_{ice}=Q_w+Q_{Al}$
We plug in the known values to obtain:
$(0.035Kg)(33.5\times 10^4J/Kg+(4186J/Kg.C^{\circ}))=[(0.062Kg)(900J/Kg.C^{\circ})(23C^{\circ}-T_f)+(0.11Kg)(4186/Kg.C^{\circ})(23C^{\circ}-T_f)]$
This simplifies to:
$T_f=0.22C^{\circ}$
(b) We know that silver has a smaller specific heat than aluminum; hence, the equilibrium temperature with silver is less than that with aluminum.
