Question #224808
  1. Find the steady state temperature of the black body disc having radius 2cm and mass 7g at 450C surrounding temperature, when kept initially at 280C. Hence determine Stefan's constant.
1
Expert's answer
2021-08-10T17:55:16-0400

Solution:-

The disc gains heat QinQ_{in} due to thermal conductivity from the surrounding and losses heat QoutQ_{out} due to radiation .

Thus, according to the Stefan Boltzmann law

Losses rate

dQoutdt=πr2σ(Tsur3T4)σ=5.56×108W/m2K4\frac{dQ_{out}}{dt}=\pi r^2\sigma(T_{sur}^3-T^4)\\\sigma=5.56\times10^{-8}W/m^2K^4\\


dQoutdt=ddt(mc(TTin))=mcdTdt\frac{dQ_{out}}{dt}=\frac{d}{dt}(mc(T-T_{in}))=mc\frac{dT}{dt}

Steady state

dQoutdt=dQindt\frac{dQ_{out}}{dt}=\frac{dQ_{in}}{dt}

dTdt=πr2σTsur4mc\frac{dT}{dt}=\frac{\pi r^2\sigma T_{sur}^4}{mc} =πr2σT4mc=\frac{\pi r^2\sigma T^4}{mc}

Solution thi equation

T(t)=13πr2σtmc+1(TstartTsur)3+TsurT(t)=\frac{1}{3\sqrt{{\frac{\pi r^2\sigma t}{mc}}+\frac{1}{(T_{start}-T_{sur})^3}}}+T_{sur}

t=infinite

Put value

T(t)=TsurT(t)=T_{sur}


Tsteady=Tsur=273+45=318K=45°CT_{steady}=T_{sur}=273+45=318K=45°C


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