## Related questions with answers

What maximum power can be radiated by a 10-cm-diameter solid lead sphere? Assume an emissivity of 1.

Solution

VerifiedThe power $P:=\frac{dQ}{dt}$ that is radiated from an object with surface $A$, that has emissivity $e$, at temperature $T$, is given by

$P=e\sigma AT^4,$

where $\sigma$=5.67$\cdot 10^{-8}~\mathrm{\frac{W}{m^2K^4}}$ is the $\textit{Stefan-Boltzmann constant}$.

At first it might seem there is a problem, since we're not given a definite temperature: that'd mean the answer is infinity, since we can increase the temperature to infinity. However, we are told we have a $\text{\underline{solid lead sphere}}$, which means that we can increase the temperature only to the melting point of lead, which is $T=328^{\circ}~\mathrm{C}$=601 K.

Having said this, we now need to substitute the area of the sphere: for a diameter $D$, the area of a sphere is

$A=\pi D^2$

Substituting, we get the maximum radiated power to be

$P=e\sigma \pi D^2T^4$

Substituting the numerical values, we can find

$P=1\cdot 5.67\cdot 10^{-8}\cdot \pi \cdot 0.1^2\cdot 601^4=\boxed{232~\mathrm{W}}$

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