Let S be the specific heat of aluminium
Q given = Q used
⇒1000×0.2×S×(150−40)=150×1×(40−27)+25×1×(40−27)
⇒200×S×110=150×13+25×13
\Rightarrow S=\frac{13\times 175}{200\times 110}=0.1034 \mathrm{cal}{g}^{-1} ^{\circ}C
\Rightarrow S=103.4 \times 4.2 J{\mathrm{kg}}^{-1}^{\circ}C=434J{\mathrm{kg}}^{-1}^{\circ}C