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(C) The rate of heat loss by the sphere is given by the Stefan-Boltzmann law: $dQ/dt = \sigma A (T^4 - T_0^4)$,where $A = 4\pi R^2$.Also,the heat lost

Vedclass · Accepted Answer

$\frac{M\alpha}{16\pi R^2\sigma} \ln \left( \frac{16}{3} \right)$

Vedclass · Answer

(C) The rate of heat loss by the sphere is given by the Stefan-Boltzmann law: $dQ/dt = \sigma A (T^4 - T_0^4)$,where $A = 4\pi R^2$.Also,the heat lost by the sphere is $dQ = -Mc dT$,where $c = \alpha T^3$.Equating the two expressions: $-M(\alpha T^3) dT = \sigma (4\pi R^2) (T^4 - T_0^4) dt$.Rearranging to solve for $dt$: $dt = -\frac{M\alpha T^3 dT}{\sigma (4\pi R^2) (T^4 - T_0^4)}$.Integrating from $T = 3T_0$ to $T = 2T_0$: $t = \int_{2T_0}^{3T_0} \frac{M\alpha T^3}{4\pi R^2 \sigma (T^4 - T_0^4)} dT$.Let $u = T^4 - T_0^4$,then $du = 4T^3 dT$,so $T^3 dT = du/4$.$t = \frac{M\alpha}{4\pi R^2 \sigma} \int_{T=2T_0}^{T=3T_0} \frac{du/4}{u} = \frac{M\alpha}{16\pi R^2 \sigma} [\ln(u)]_{T=2T_0}^{T=3T_0}$.$t = \frac{M\alpha}{16\pi R^2 \sigma} [\ln(T^4 - T_0^4)]_{2T_0}^{3T_0} = \frac{M\alpha}{16\pi R^2 \sigma} \ln \left( \frac{(3T_0)^4 - T_0^4}{(2T_0)^4 - T_0^4} \right)$.$t = \frac{M\alpha}{16\pi R^2 \sigma} \ln \left( \frac{81T_0^4 - T_0^4}{16T_0^4 - T_0^4} \right) = \frac{M\alpha}{16\pi R^2 \sigma} \ln \left( \frac{80}{15} \right) = \frac{M\alpha}{16\pi R^2 \sigma} \ln \left( \frac{16}{3} \right)$.

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