Question

For a black body at temperature$727\,{}^\circ C$, its radiating power is 60 watt and temperature of surrounding is $227\,{}^\circ C$. If the temperature of the black body is changed to $1227\,{}^\circ C$then its radiating power will be:-
A. 304 W
B. 320 W
C. 240 W
D. 120 W

Answer 1

We will make use of the formula that is used for computing the thermal power of radiation. We will consider two situations and will find the ratio of the same. Then we will substitute the given values in the equation obtained to find the value of the radiating power of the black body.

Formula used:
$\dfrac{dQ}{dt}=\sigma Ae{{T}^{4}}$

Complete step by step solution:
The formula used is called the thermal power of radiation and is given as follows.
$\dfrac{dQ}{dt}=\sigma Ae{{T}^{4}}$
Where $\sigma$is the Stefan – Boltzmann constant and e is the emissivity
The emissivity is the fraction of the incoming radiation that is absorbed. The emissivity of 1 corresponds to the black body.
The thermal power of radiation for the temperature values of $227\,{}^\circ C$ and $727\,{}^\circ C$.
Firstly, we will convert the unit of temperature from degree Celsius to Kelvin.
$227\,{}^\circ C$ = 500 K
$727\,{}^\circ C$ = 1000 K
${{E}_{1}}=\sigma Ae\left[ {{\left( {{T}_{1}} \right)}^{4}}-{{({{T}_{2}})}^{4}} \right]$
Substitute the values of the temperature and the radiating power in the above equation.
$60=\sigma Ae\left[ {{\left( 1000 \right)}^{4}}-{{(500)}^{4}} \right]$…… (1)
The thermal power of radiation for the temperature values of $227\,{}^\circ C$ and $1227\,{}^\circ C$.
Firstly, we will convert the unit of temperature from degree Celsius to Kelvin.
$227\,{}^\circ C$ = 500 K
$1227\,{}^\circ C$ = 1500 K
${{E}_{2}}=\sigma Ae\left[ {{\left( {{T}_{1}} \right)}^{4}}-{{({{T}_{2}})}^{4}} \right]$
Substitute the values of the temperature and the radiating power in the above equation.
$E=\sigma Ae\left[ {{\left( 1500 \right)}^{4}}-{{(500)}^{4}} \right]$…… (2)
Divide the equations (1) and (2)
$\dfrac{60}{E}=\dfrac{\sigma Ae\left[ {{\left( 1000 \right)}^{4}}-{{(500)}^{4}} \right]}{\sigma Ae\left[ {{\left( 1500 \right)}^{4}}-{{(500)}^{4}} \right]}$
Continue the further calculation

& \dfrac{60}{E}=\dfrac{{{500}^{4}}({{2}^{4}}-1)}{{{500}^{4}}({{3}^{4}}-1)} \\\ & E=\dfrac{60\times 80}{15} \\\ & \Rightarrow E=320\,W \\\ \end{aligned}$$ $$\therefore $$ The value of the thermal radiation of the black body is 320 W. **As the value of the thermal radiation of the black body is 320 W, thus, the option (B) is correct.** **Note:** The formula of the thermal radiation consists of only one temperature value, but we have converted it into such that we can substitute the two values, so this is the important step in the calculation part.