A black coloured solid sphere of radius R and mass M is inside a cavity with a vacuum inside. The walls of the cavity are maintained at temperature T0. The initial temperature of the sphere is 3T0. If the specific heat o

Step-by-step solution

1. 1

   Given parameters

   Sphere radius $R$, mass $M$, initial temperature $3T_0$, final temperature $2T_0$, cavity temperature $T_0$. Specific heat $c = \alpha T^3$. Stefan-Boltzmann constant $\sigma$.
2. 2

   Rate of heat loss

   Rate of heat loss by radiation from a black body is given by Stefan-Boltzmann Law.

   $$dQ/dt = A\sigma(T^4 - T_0^4)$$
3. 3

   Relate heat loss to temperature change

   Heat lost by sphere $dQ = mc dT$. Equate this with radiation heat loss and integrate.

   $$mc dT = -A\sigma(T^4 - T_0^4) dt$$
4. 4

   Substitute values

   Substitute $A = 4\pi R^2$ and $c = \alpha T^3$ into the equation.

   $$M(\alpha T^3) dT = -4\pi R^2\sigma(T^4 - T_0^4) dt$$
5. 5

   Rearrange for integration

   Separate variables $T$ and $t$ for integration.

   $$dt = -\frac{M\alpha T^3}{4\pi R^2\sigma(T^4 - T_0^4)} dT$$
6. 6

   Integrate both sides

   Integrate $t$ from $0$ to $\tau$ and $T$ from $3T_0$ to $2T_0$.

   $$\int_0^\tau dt = -\frac{M\alpha}{4\pi R^2\sigma} \int_{3T_0}^{2T_0} \frac{T^3}{T^4 - T_0^4} dT$$
7. 7

   Solve the integral

   Use substitution $u = T^4 - T_0^4$, so $du = 4T^3 dT$.

   $$\int \frac{T^3}{T^4 - T_0^4} dT = \frac{1}{4} \ln|T^4 - T_0^4|$$
8. 8

   Evaluate definite integral

   Substitute limits and simplify to find $\tau$.

   $$\tau = \frac{M\alpha}{16\pi R^2\sigma} [\ln(T^4 - T_0^4)]_{2T_0}^{3T_0}$$
9. 9

   Final time $\tau$

   Simplify the logarithmic terms.

   \tau = \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{80T_0^4}{15T_0^4}\right) = \frac{M\alpha}{16\pi R^2\sigm