Answer
the net heat transfer through the black body disc is given by
dQindt=mcdTdt\frac{dQ_{in}}{dt}=mc\frac{dT}{dt}dtdQin=mcdtdT
So heat transffer per unit second
dQ=mc dTdQ=mc\space dTdQ=mc dT
Using given data
dQ=0.009×0.385×(330+273)=2.09JdQ=0.009\times0.385\times(330+273) \\=2.09JdQ=0.009×0.385×(330+273)=2.09J
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