Dane:

$T=4000\text{K}$

$\Delta T=800\text{K}$

Rozwiązanie:

a)

Dla ciała doskonale czarnego natężenie $I$   promieniowania przez nie emitowanego wyrażamy jako:

$I=\sigma T^4$  

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$\sigma =5,67\cdot {10}^{-8}\frac{\text{W}}{{\text{m}}^2\cdot {\text{K}}^4}-\text{stała Stefana-Boltzmanna}$  

$T-\text{temperatura ciała}$  

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Natężenie promieniowania możemy wyrazić jako:

$I=\frac{P}{S}$  

$P-\text{moc promieniowania ciała}$

$S-\text{pole powierzchni ciała}$

 

Zatem:

$\frac{P}{S}=\sigma T^4$  

$P=\sigma S{T}^4$  

 

Początkowa moc promieniowania ciała jest równa:

$P_1=\sigma S{T}^4$  

 

Po zmianie temperatury ciała jego moc promieniowania wyrazimy jako sumę mocy promieniowania obu połów tego ciała.

$P_2=P_A+P_B$  

$P_A=\sigma \frac{S}{2}{T}_A^4=\frac{1}{2}\sigma S{\left(T+\Delta T\right)}^4$  

$P_B=\sigma \frac{S}{2}{T}_B^4=\frac{1}{2}\sigma S{\left(T-\Delta T\right)}^4$  

Zatem:

$P_2=\frac{1}{2}\sigma S{\left(T+\Delta T\right)}^4+\frac{1}{2}\sigma S{\left(T-\Delta T\right)}^4=\frac{1}{2}\sigma S\left({\left(T+\Delta T\right)}^4+{\left(T-\Delta T\right)}^4\right)$  

 

Obliczmy szukany stosunek mocy promieniowania tego ciała:

$\frac{P_2}{P_1}=\frac{\frac{1}{2}\sigma S\left({\left(T+\Delta T\right)}^4+{\left(T-\Delta T\right)}^4\right)}{\sigma S{T}^4}$  

$\frac{P_2}{P_1}=\frac{\frac{1}{2}\left({\left(T+\Delta T\right)}^4+{\left(T-\Delta T\right)}^4\right)}{T^4}$  

$\frac{P_2}{P_1}=\frac{{\left(T+\Delta T\right)}^4+{\left(T-\Delta T\right)}^4}{2T^4}$  

$\frac{P_2}{P_1}=\frac{{\left(4000\text{K}+800\text{K}\right)}^4+{\left(4000\text{K}-800\text{K}\right)}^4}{2\cdot {\left(4000\text{K}\right)}^4}=\frac{{\left(4800\text{K}\right)}^4+{\left(3200\text{K}\right)}^4}{2\cdot {\left(4000\text{K}\right)}^4}=\frac{{\left(4,8\cdot {10}^3\text{K}\right)}^4+{\left(3,2\cdot {10}^3\text{K}\right)}^4}{2\cdot {\left(4\cdot {10}^3\text{K}\right)}^4}\approx \frac{530,8\cdot {10}^{12}\text{K}+104,9\cdot {10}^{12}\text{K}}{512\cdot {10}^{12}\text{K}}=\frac{635,7\cdot {10}^{12}\text{K}}{512\cdot {10}^{12}\text{K}}\approx 1,24$  

$\frac{P_2}{P_1}\approx 124\%$  

 

b)

Dla dowolnego podziału powierzchni ciała zapiszemy:

$P_A=\sigma \left(S-S_1\right)T_A^4=\sigma \left(S-S_1\right){\left(T+\Delta T\right)}^4$  

$P_B=\sigma S_1{T}_B^4=\sigma S_1{\left(T-\Delta T\right)}^4$  

$P_2=P_A+P_B$

$P_2=\sigma \left(S-S_1\right){\left(T+\Delta T\right)}^4+\sigma S_1{\left(T-\Delta T\right)}^4=\sigma \left(\left(S-S_1\right){\left(T+\Delta T\right)}^4+S_1{\left(T-\Delta T\right)}^4\right)$   

 

Moc początkowa ciała pozostaje bez zmian:

$P_1=\sigma S{T}^4$  

 

Chcemy, aby w obu przypadkach moce promieniowania ciała były takie same:

$P_1=P_2$  

$\sigma S{T}^4=\sigma \left(\left(S-S_1\right){\left(T+\Delta T\right)}^4+S_1{\left(T-\Delta T\right)}^4\right)\mid :\sigma$  

$S{T}^4=\left(S-S_1\right){\left(T+\Delta T\right)}^4+S_1{\left(T-\Delta T\right)}^4\mid :S$  

$T^4=\frac{S-S_1}{S}{\left(T+\Delta T\right)}^4+\frac{S_1}{S}{\left(T-\Delta T\right)}^4$  

$T^4=\left(1-\frac{S_1}{S}\right){\left(T+\Delta T\right)}^4+\frac{S_1}{S}{\left(T-\Delta T\right)}^4$  

$T^4={\left(T+\Delta T\right)}^4-\frac{S_1}{S}{\left(T+\Delta T\right)}^4+\frac{S_1}{S}{\left(T-\Delta T\right)}^4$  

$-{\left(T+\Delta T\right)}^4+T^4=-\frac{S_1}{S}{\left(T+\Delta T\right)}^4+\frac{S_1}{S}{\left(T-\Delta T\right)}^4\mid \cdot \left(-1\right)$  

${\left(T+\Delta T\right)}^4-T^4=\frac{S_1}{S}{\left(T+\Delta T\right)}^4-\frac{S_1}{S}{\left(T-\Delta T\right)}^4$  

${\left(T+\Delta T\right)}^4-T^4=\frac{S_1}{S}\left({\left(T+\Delta T\right)}^4-{\left(T-\Delta T\right)}^4\right)$  

$\frac{S_1}{S}=\frac{{\left(T+\Delta T\right)}^4-T^4}{{\left(T+\Delta T\right)}^4-{\left(T-\Delta T\right)}^4}$  

Upraszczamy otrzymany wzór:

$\frac{S_1}{S}=\frac{\frac{{\left(T+\Delta T\right)}^4}{{\left(T+\Delta T\right)}^4}-\frac{T^4}{{\left(T+\Delta T\right)}^4}}{\frac{{\left(T+\Delta T\right)}^4}{{\left(T+\Delta T\right)}^4}-\frac{{\left(T-\Delta T\right)}^4}{{\left(T+\Delta T\right)}^4}}$  

$\frac{S_1}{S}=\frac{1-\frac{T^4}{{\left(T+\Delta T\right)}^4}}{1-\frac{{\left(T-\Delta T\right)}^4}{{\left(T+\Delta T\right)}^4}}$  

$\frac{S_1}{S}=\frac{1-{\left(\frac{T}{T+\Delta T}\right)}^4}{1-{\left(\frac{T-\Delta T}{T+\Delta T}\right)}^4}$  

Obliczmy stosunek zadanych powierzchni:

$\frac{S_1}{S}=\frac{1-{\left(\frac{4000\text{K}}{4000\text{K}+800\text{K}}\right)}^4}{1-{\left(\frac{4000\text{K}-800\text{K}}{4000\text{K}+800\text{K}}\right)}^4}\approx \frac{1-{\left(0,833\right)}^4}{1-{\left(0,667\right)}^4}\approx \frac{1-0,481}{1-0,198}=\frac{0,519}{0,802}\approx 0,647$  

$S_1\approx 64,7\%S$