Heat loss = Heat gain =mSΔθ So,mASAΔθA=mBSBΔθB $\begin{array}{l}
\Rightarrow 100 \times \mathrm{S}{\mathrm{A}} \times(100-90)=50 \times \mathrm{S}{\mathrm{B}} \times(90-75) \
2 \mathrm{S}_{\mathrm{A}}=1.5 \mathrm{S}{\mathrm{B}} \Rightarrow \mathrm{S}{\mathrm{A}}=\frac{3}{4} \mathrm{~S}_{\mathrm{B}}
\end{array}Now,100 \times \mathrm{S}{\mathrm{A}} \times(100-\theta)=50 \times \mathrm{S}{\mathrm{B}} \times(\theta-50)2 \times\left(\frac{3}{4}\right) \times(100-\theta)=(\theta-50)\begin{array}{l}
300-3 \theta=2 \theta-100 \
400=5 \theta \Rightarrow \theta=80^{\circ} \mathrm{C}
\end{array}$