Question

Calculate the energy radiated in one minute by a black body of surface area $100c{m^2}$ when it is maintained at $227^\circ C$ . ( $\sigma = 5.67 \times {10^{ - 8}}J/{m^2}s{K^4}$ )

Answer 1

Since the body in consideration is black, its emissivity will be 1( $\varepsilon = 1$ ). We can use Stefan-Boltzmann law to find the power dissipated by the black body and further calculate the energy radiated by multiplying the power with the time given.

Complete step-by-step solution:
According to Stefan-Boltzmann law, power radiated from an object of surface area A,
$P = A\varepsilon \sigma {T^4}$
Absolute temperature = temperature in degree Celsius + 273
$\Rightarrow T = 227^\circ C + 273 = 500K$
Substituting the values of surface area, emissivity and temperature in the expression for power, we get $P = (100 \times {10^{ - 4}}) \times 1 \times (5.67 \times {10^{ - 8}}) \times ({500^{^{_4}}}) = 35.43J/s$
Now we know that energy = power * time
**Hence, Energy radiated in 1 minute (60 seconds), $E = 35.43J/s \times 60s = 2126J$.
**
Additional Information:
A cooler body radiates less energy than a warmer body. The Stefan-Boltzmann law describes the power dissipated from a black body in terms of its absolute temperature T. The constant of proportionality called the Stefan-Boltzmann constant, $\sigma = \dfrac{{2{\pi ^5}{k^4}}}{{15{c^2}{h^3}}}$
Both Stefan’s law and Wein’s displacement law were confirmed by measuring the black body curves at different temperatures.
Another expression for power radiated by a black body is $\vartriangle P = \dfrac{{2\pi h{c^2}}}{{{\lambda ^5}}}\dfrac{1}{{{e^{\dfrac{{hc}}{{\lambda kt}}}} - 1}}\Delta \lambda A$.

Note:- Another alternative method of calculating power radiated by a black body is through the graph of power density against wavelength. The numerical sum of values of Planck radiation density times a wavelength interval is taken and divided into 100 parts and their brute force term is taken. But you need not go into that much detail at the moment.