The total energy radiated from a black body s | vedclass

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Question

(D) According to the Stefan-Boltzmann law,the total energy $Q$ radiated from a black body per unit time is proportional to the fourth power of its abs

Vedclass · Accepted Answer

$28$

Vedclass · Answer

(D) According to the Stefan-Boltzmann law,the total energy $Q$ radiated from a black body per unit time is proportional to the fourth power of its absolute temperature $T$,i.e.,$Q \propto T^4$.Let $Q_1$ be the energy radiated at temperature $T$ and $Q_2$ be the energy radiated at temperature $2T$.Then,$\frac{Q_2}{Q_1} = \left( \frac{2T}{T} ight)^4 = 2^4 = 16$.So,$Q_2 = 16 Q_1$.The heat absorbed by water is given by $Q = ms\Delta	heta$,where $m$ is the mass of water and $s$ is the specific heat capacity.For the first case: $Q_1 = ms(20.5 - 20) = ms(0.5)$.For the second case: $Q_2 = ms(	heta - 20)$,where $	heta$ is the final temperature.Dividing the two equations:$\frac{Q_2}{Q_1} = \frac{ms(	heta - 20)}{ms(0.5)} = \frac{	heta - 20}{0.5}$.Since $\frac{Q_2}{Q_1} = 16$,we have:$16 = \frac{	heta - 20}{0.5}$.$8 = 	heta - 20$.$	heta = 28^{\circ}C$.

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