Answer
$\Delta S_1=-22.1\frac{J}{K}$
Work Step by Step
We can find entropy change of aluminium as:
$\Delta S_1=\int _{Q_i}^{Q_f}\frac{dQ_1}{T}=\int _{Q_i}^{Q_f} \frac{m_1c_1 dT}{T}=m_1c_1ln \frac{T}{T_{i1}}$
We plug in the known values to obtain:
$\Delta S_1=0.2(900)\ln \frac{330}{373}=-22.1\frac{J}{K}$
