{L}_{Cu}=46\mathrm{cm};{L}_{B}=13\mathrm{cm}&{L}_{S}=12\mathrm{cm}.

\begin{matrix}{K}_{Cu}=0.92 \\ {K}_{B}=0.26 \\ {K}_{S}=0.12\end{matrix}}CGSunit
We know that R=KAL
RCu=0.92×446=12.5
RB=0.26×413=12.5
RS=0.12×412=25
Now, if T is temperature of junction then
RCu100−T=RBT−0+RST−0
12.5100−T=12.5T+25T
∴12.5100−T=253T
200=5T
∴T=40∘
⇒QCu=RCu100−T
=12.5100−40
=12.560
⇒QCu=4.8Cal/s