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Q. The original temperature of a black body is $727^{\circ}C $ . The temperature at which this black body must be raised so as to double, the total radiant energy, is

Punjab PMETPunjab PMET 2001Thermal Properties of Matter

Solution:

Using the Stefan's law, the total radiant energy is
$\theta=e A \sigma T^{4} t \propto T^{4}$
hence $\frac{E_{2}}{E_{1}}=\frac{T_{2}^{4}}{T_{1}^{4}}$
$=\left(\frac{T_{2}}{T_{1}}\right)^{4}$
here $\frac{E_{2}}{E_{1}}=2$
$T_{1} =727^{\circ} C +273=1000\, K$
$2^{1 / 4} =\frac{T_{2}}{1000}$
or $T_{2}=1000 \times 2^{1 / 4}=1000 \times 1.19$
$=1190\, K$