For a black body at temperature $727^{o}$ its rate of energy loss is $20$ watt and temperature of surrounding is $227^{o}$ C. If temperature of black body is changed to $1227^{o}$ C then its rate of energy loss will be

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Solution

According to stephen Boltzman Law ,
Heat loss by a black body $(E) α (T^{4} - {T_{0}}^{4})$
or $\frac{E_{1}}{E_{2}} = \frac{(T_{1}^{4} - T_{0}^{4})}{(T_{2}^{4} - T_{0}^{4})}$ ...... (A)
$T_{1}$ = temperature of the body
$T_{0}$ =temperature surrounding
Given, $T_{1} = 727^{0} c, T_{2} = 1227^{0} c, T_{0} = 227^{0} c$
$E_{1} = 20 w a t t$
Putting the above values in A, we get,
$E_{2} = \frac{E_{1} (T_{2}^{4} - T_{0}^{4})}{((T_{1}^{4} - T_{0}^{4})}$
= $= 20 \frac{(1227^{4} - 227^{4})}{(727^{4} - 227^{4})} = 166.40 w$

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