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(A) According to the Stefan-Boltzmann law,the power radiated by a surface is given by $P = \sigma A e T^4$,where $\sigma$ is the Stefan-Boltzmann cons

Vedclass · Accepted Answer

$9$

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(A) According to the Stefan-Boltzmann law,the power radiated by a surface is given by $P = \sigma A e T^4$,where $\sigma$ is the Stefan-Boltzmann constant,$A$ is the surface area,$e$ is the emissivity,and $T$ is the absolute temperature.Since $A$,$e$,and $\sigma$ are constant,we can write $P = k T^4$ for some constant $k$.The sensor displays $S = \log_{2}(P / P_0) = \log_{2}(k T^4 / P_0)$.Let $C = k / P_0$,then $S = \log_{2}(C T^4) = \log_{2}(C) + 4 \log_{2}(T)$.At $T_1 = 487^{\circ} C = 487 + 273 = 760 \ K$,the sensor reading is $S_1 = 1$.So,$1 = \log_{2}(C) + 4 \log_{2}(760)$.At $T_2 = 2767^{\circ} C = 2767 + 273 = 3040 \ K$,the sensor reading is $S_2 = \log_{2}(C) + 4 \log_{2}(3040)$.Subtracting the two equations: $S_2 - 1 = 4 \log_{2}(3040) - 4 \log_{2}(760) = 4 \log_{2}(3040 / 760)$.Since $3040 / 760 = 4$,we have $S_2 - 1 = 4 \log_{2}(4) = 4 	imes 2 = 8$.Therefore,$S_2 = 8 + 1 = 9$.

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