A copper sphere has radius of 5cm. Its outer surface is coated black. Time required for the black to cool down from 1000K to 300K is _______ ×103s? (density of copper = 9 x 103 kg/m3 and specific heat = 4 kJ/kg)(Round off to nearest integer)

dQdt=-eσAT4 ; dQdT=mcFrom the above equations,∫0tdt=−mceσA∫T1T2T−4dT=mc3eσA(1T23−1T13) where mA=rρ3 for sphereUse e=1.t=rρc9σ(T13−T23T13T23)=5×10−2×9×103×4×1039×5.67×10−8(10003−30033×109)=127×103s

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A copper sphere has radius of 5cm. Its outer surface is coated black. Time required for the black to cool down from 1000K to 300K is _______ $\times 10^{3}$ s? (density of copper = 9 x 10³ kg/m³ and specific heat = 4 kJ/kg)
(Round off to nearest integer)

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answer is 127.

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Detailed Solution

$\frac{d Q}{d t} = - e σ A T^{4}; \frac{d Q}{d T} = m c$

From the above equations,

$\int_{0}^{t} d t = - \frac{m c}{e σ A} \int_{T_{1}}^{T_{2}} T^{- 4} d T = \frac{m c}{3 e σ A} (\frac{1}{T_{2}^{3}} - \frac{1}{T_{1}^{3}}) w h e r e \frac{m}{A} = \frac{r ρ}{3} f o r s p h e r e$

Use e=1.

$t = \frac{r ρ c}{9 σ} (\frac{T_{1}^{3} - T_{2}^{3}}{T_{1}^{3} T_{2}^{3}}) = \frac{5 \times 10^{- 2} \times 9 \times 10^{3} \times 4 \times 10^{3}}{9 \times 5.67 \times 10^{- 8}} (\frac{1000^{3} - 300^{3}}{3 \times 10^{9}}) = 127 \times 10^{3} s$