Question

If the temperature of the hot body is raised by $5\%$ the rate of heat radiated would be increased by how much percentage?
(A) $12\%$
(B) $22\%$
(C) $32\%$
(D) $42\%$

Answer 1

The heat energy radiated is directly proportional to the fourth power of the temperature of the black body. The percentage increase is the difference between the new value and the old value divided by the new value.
Formula used: In this solution we will be using the following formulae;
$H = \sigma A{T^4}$ where $H$ is the heat energy radiated, $\sigma$ is the Stefan Boltzmann constant, $A$ is area of the surface of the blackbody, and $T$ is the absolute temperature of the black body.
$PI = \dfrac{{NV - OV}}{{OV}}$ where $PI$ is the percentage increase of a particular value, $NV$ is the new value, and $OV$ is the old value.

Complete Step-by-Step solution:
Generally, the heat energy radiated by a black body is directly related to the fourth power of the temperature of that black body as given by the Stefan’s law as
$H = \sigma A{T^4}$ where $\sigma$ is the Stefan Boltzmann constant, $A$ is area of the surface of the blackbody, and $T$ is the absolute temperature of the black body
Temperature increasing by 5 percent signifies the final temperature to be
$T' = T + \dfrac{5}{{100}}T$ which by adding and simplifying gives,
$T' = \dfrac{{21}}{{20}}T$
$\Rightarrow \dfrac{{T'}}{T} = \dfrac{{21}}{{20}}$
Percentage error can be defined as
$PI = \dfrac{{NV - OV}}{{OV}}$ where $PI$ is the percentage increase of a particular value, $NV$ is the new value, and $OV$ is the old value.
Hence, percentage increase in the heat energy radiated would be defined as
$PI = \dfrac{{H' - H}}{H} \times 100\%$
Where
$H' = \sigma AT{'^4}$
Hence,
$\dfrac{{H'}}{H} = \dfrac{{\sigma AT{'^4}}}{{\sigma A{T^4}}} = \dfrac{{T{'^4}}}{{{T^4}}} = {\left( {\dfrac{{T'}}{T}} \right)^4}$
By inserting known values, we have
$\dfrac{{H'}}{H} = {\left( {\dfrac{{21}}{{20}}} \right)^4}$
Hence, by multiplying both sides by $H$, we get
$H' = {\left( {\dfrac{{21}}{{20}}} \right)^4}H$
Going back to the definition, and inserting the value above into it we have
$PI = \dfrac{{{{\left( {\dfrac{{21}}{{20}}} \right)}^4}H - H}}{H} \times 100\%$
Dividing numerator and denominator by $H$, we get
$PI = \left[ {{{\left( {\dfrac{{21}}{{20}}} \right)}^4} - 1} \right] \times 100\%$
$PI = \left[ {{{\left( {1.05} \right)}^4} - 1} \right] \times 100\%$
Hence, finding the fourth power, we get
$PI = \left[ {1.22 - 1} \right] \times 100\%$
Computing the above relation, we get
$PI = 22\%$

Hence, the correct option is B.

Note: For clarity; you might have seen a text where heat radiated from a blackbody is written as
$H = \varepsilon \sigma A{T^4}$ where $\varepsilon$ is the emissivity of the body. This equation and the one above are identical, this is because for a black body, the emissivity is equal to 1, and hence can drop out of the equation. This equation is more generally used for heat radiated by any type of body.