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Question Number 10807 by Saham last updated on 25/Feb/17

Commented by ridwan balatif last updated on 26/Feb/17

19.13  ε=1(because like act blackbody), A=1cm^2 =10^(−4) m^2                  σ(Boltzman constant)=5.67×10^(−8) Wm^(−2) K^(−4) , T=1727^o C=2000K                       (Q/t)=ε×σ×A×T^4                        (Q/t)=1×5.67×10^(−8) ×1×10^(−4) ×(2000)^4                        (Q/t)=90.72J.s^(−1) =21.68cal/s  Question 19.14 have the same way

$$\mathrm{19}.\mathrm{13}\:\:\epsilon=\mathrm{1}\left(\mathrm{because}\:\mathrm{like}\:\mathrm{act}\:\mathrm{blackbody}\right),\:\mathrm{A}=\mathrm{1cm}^{\mathrm{2}} =\mathrm{10}^{−\mathrm{4}} \mathrm{m}^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\sigma\left(\mathrm{Boltzman}\:\mathrm{constant}\right)=\mathrm{5}.\mathrm{67}×\mathrm{10}^{−\mathrm{8}} \mathrm{Wm}^{−\mathrm{2}} \mathrm{K}^{−\mathrm{4}} ,\:\mathrm{T}=\mathrm{1727}^{\mathrm{o}} \mathrm{C}=\mathrm{2000K} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\frac{\mathrm{Q}}{\mathrm{t}}=\epsilon×\sigma×\mathrm{A}×\mathrm{T}^{\mathrm{4}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\frac{\mathrm{Q}}{\mathrm{t}}=\mathrm{1}×\mathrm{5}.\mathrm{67}×\mathrm{10}^{−\mathrm{8}} ×\mathrm{1}×\mathrm{10}^{−\mathrm{4}} ×\left(\mathrm{2000}\right)^{\mathrm{4}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\frac{\mathrm{Q}}{\mathrm{t}}=\mathrm{90}.\mathrm{72J}.\mathrm{s}^{−\mathrm{1}} =\mathrm{21}.\mathrm{68cal}/\mathrm{s} \\ $$$$\mathrm{Question}\:\mathrm{19}.\mathrm{14}\:\mathrm{have}\:\mathrm{the}\:\mathrm{same}\:\mathrm{way} \\ $$

Commented by Saham last updated on 26/Feb/17

God bless you sir.

$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}. \\ $$

Answered by sandy_suhendra last updated on 26/Feb/17

19.12  A=0.8 m×0.8 m=0.64 m^2   d=0.3 cm=3×10^(−3)  m  ΔT=20 K  k_1 =k_3 =0.84 W/Km  k_2 =0.08 W/Km  (1/k_(total) )=(1/k_1 )+(1/k_2 )+(1/k_3 )            =(1/(0.84))+(1/(0.08))+(1/(0.84))=(1/(0.0672))  k_(total) =0.0672 W/Km    (Q/t)=((k_(total) .A.ΔT)/d)=((0.0672×0.64×20)/(3×10^(−3) ))       =286.72 J/s ×0.24       =68.8 cal/s ≈ 69 cal/s  ( 1Joule = 0.24 cal )

$$\mathrm{19}.\mathrm{12} \\ $$$$\mathrm{A}=\mathrm{0}.\mathrm{8}\:\mathrm{m}×\mathrm{0}.\mathrm{8}\:\mathrm{m}=\mathrm{0}.\mathrm{64}\:\mathrm{m}^{\mathrm{2}} \\ $$$$\mathrm{d}=\mathrm{0}.\mathrm{3}\:\mathrm{cm}=\mathrm{3}×\mathrm{10}^{−\mathrm{3}} \:\mathrm{m} \\ $$$$\Delta\mathrm{T}=\mathrm{20}\:\mathrm{K} \\ $$$$\mathrm{k}_{\mathrm{1}} =\mathrm{k}_{\mathrm{3}} =\mathrm{0}.\mathrm{84}\:\mathrm{W}/\mathrm{Km} \\ $$$$\mathrm{k}_{\mathrm{2}} =\mathrm{0}.\mathrm{08}\:\mathrm{W}/\mathrm{Km} \\ $$$$\frac{\mathrm{1}}{\mathrm{k}_{\mathrm{total}} }=\frac{\mathrm{1}}{\mathrm{k}_{\mathrm{1}} }+\frac{\mathrm{1}}{\mathrm{k}_{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{k}_{\mathrm{3}} } \\ $$$$\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{1}}{\mathrm{0}.\mathrm{84}}+\frac{\mathrm{1}}{\mathrm{0}.\mathrm{08}}+\frac{\mathrm{1}}{\mathrm{0}.\mathrm{84}}=\frac{\mathrm{1}}{\mathrm{0}.\mathrm{0672}} \\ $$$$\mathrm{k}_{\mathrm{total}} =\mathrm{0}.\mathrm{0672}\:\mathrm{W}/\mathrm{Km} \\ $$$$ \\ $$$$\frac{\mathrm{Q}}{\mathrm{t}}=\frac{\mathrm{k}_{\mathrm{total}} .\mathrm{A}.\Delta\mathrm{T}}{\mathrm{d}}=\frac{\mathrm{0}.\mathrm{0672}×\mathrm{0}.\mathrm{64}×\mathrm{20}}{\mathrm{3}×\mathrm{10}^{−\mathrm{3}} } \\ $$$$\:\:\:\:\:=\mathrm{286}.\mathrm{72}\:\mathrm{J}/\mathrm{s}\:×\mathrm{0}.\mathrm{24} \\ $$$$\:\:\:\:\:=\mathrm{68}.\mathrm{8}\:\mathrm{cal}/\mathrm{s}\:\approx\:\mathrm{69}\:\mathrm{cal}/\mathrm{s} \\ $$$$\left(\:\mathrm{1Joule}\:=\:\mathrm{0}.\mathrm{24}\:\mathrm{cal}\:\right) \\ $$

Commented by Saham last updated on 26/Feb/17

God bless you sir.

$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}. \\ $$

Answered by sandy_suhendra last updated on 26/Feb/17

19.14  e=0.83  A=50 mm^2  = 5×10^(−5)  m^2   T=2,127+273=2,400 K  (Q/t)=eσAT^4 =0.83×5.67.10^(−8) ×5.10^(−5) ×2,400^4           =78 W

$$\mathrm{19}.\mathrm{14} \\ $$$$\mathrm{e}=\mathrm{0}.\mathrm{83} \\ $$$$\mathrm{A}=\mathrm{50}\:\mathrm{mm}^{\mathrm{2}} \:=\:\mathrm{5}×\mathrm{10}^{−\mathrm{5}} \:\mathrm{m}^{\mathrm{2}} \\ $$$$\mathrm{T}=\mathrm{2},\mathrm{127}+\mathrm{273}=\mathrm{2},\mathrm{400}\:\mathrm{K} \\ $$$$\frac{\mathrm{Q}}{\mathrm{t}}=\mathrm{e}\sigma\mathrm{AT}^{\mathrm{4}} =\mathrm{0}.\mathrm{83}×\mathrm{5}.\mathrm{67}.\mathrm{10}^{−\mathrm{8}} ×\mathrm{5}.\mathrm{10}^{−\mathrm{5}} ×\mathrm{2},\mathrm{400}^{\mathrm{4}} \:\:\: \\ $$$$\:\:\:\:\:=\mathrm{78}\:\mathrm{W} \\ $$

Commented by Saham last updated on 26/Feb/17

God bless you sir.

$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}. \\ $$

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