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(C) Let the radius of sphere $P$ be $r$ and the radius of sphere $Q$ be $3r$.The rate of heat loss by radiation is given by Stefan-Boltzmann Law: $\fr

Vedclass · Accepted Answer

$3$

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(C) Let the radius of sphere $P$ be $r$ and the radius of sphere $Q$ be $3r$.The rate of heat loss by radiation is given by Stefan-Boltzmann Law: $\frac{dQ}{dt} = \sigma e A T^4$.Also,the rate of heat loss is related to the rate of fall of temperature by the equation: $\frac{dQ}{dt} = -ms \frac{dT}{dt}$,where $m$ is mass,$s$ is specific heat capacity,and $\frac{dT}{dt}$ is the rate of fall of temperature.Equating the two: $ms \frac{dT}{dt} = \sigma e A T^4$.Since $m = \rho V = \rho (\frac{4}{3} \pi r^3)$,we have $\rho (\frac{4}{3} \pi r^3) s \frac{dT}{dt} = \sigma e (4 \pi r^2) T^4$.Thus,the rate of fall of temperature is $\frac{dT}{dt} = \frac{\sigma e (4 \pi r^2) T^4}{\rho (\frac{4}{3} \pi r^3) s} = \frac{3 \sigma e T^4}{\rho r s}$.This shows that $\frac{dT}{dt} \propto \frac{1}{r}$.Therefore,$\frac{(\frac{dT}{dt})_P}{(\frac{dT}{dt})_Q} = \frac{r_Q}{r_P} = \frac{3r}{r} = 3$.Given $(\frac{dT}{dt})_P = x (\frac{dT}{dt})_Q$,we find $x = 3$.

Spheres $P$ and $Q$ are uniformly constructed from the same material,which is a good conductor of heat,and the radius of $Q$ is thrice the radius of $P$. The rate of fall of temperature of $P$ is $x$ times that of $Q$ when both are at the same surface temperature. The value of $x$ is:

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