Courses
Courses for Kids
Free study material
Offline Centres
More
Store Icon
Store

If the temperature of a hot body is raised by 0.5%, then the heat energy radiated would increases by:
(A) 0.5%
(B) 1.0%
(C) 1.5%
(D) 2.0%

seo-qna
Last updated date: 16th Jul 2024
Total views: 62.4k
Views today: 1.62k
Answer
VerifiedVerified
62.4k+ views
Hint Heat energy radiated is proportional to fourth power of temperature. So the ratio of final heat energy to the initial is equal to the fourth power, the ratio of the final temperature to the initial temperature. The final temperature is changed by 0.5%. Incorporating this change, the change in heat radiation is evaluated.

Complete step-by-step answer
The rate at which heat energy is radiated by a hot body is given by
$Q = \sigma {T^4}A$
From this we know that
$Q \propto {T^4}$
So, let Q1 and Q2 be the initial and final heat energy at temperatures T1 and T2 respectively.
It is given that temperature is increased by 0.5%, so
T1= t K
\[{T_2} = {\text{ }}t{\text{ }} + {\text{ }}(0.5{\text{ }}\% {\text{ }} \times t){\text{ }} = 1.005t{\text{ }}K\]
Using the temperature and energy relation above,
$
\dfrac{{{Q_1}}}{{{Q_2}}} = \dfrac{{{T_1}^4}}{{{T_2}^4}} \\
{Q_2} = \dfrac{{{{(1.005t)}^4} \times {Q_1}}}{{{t^4}}} \\
{Q_2} = 1.02{Q_1} \\
$
Now, percentage increase in heat energy radiated is given by,
$
\Delta Q\% = \dfrac{{{Q_2} - {Q_1}}}{{{Q_1}}} \times 100 \\
\Delta Q\% = \dfrac{{1.02{Q_1} - {Q_1}}}{{{Q_1}}} \times 100 \\
\Delta Q\% = 0.02 \times 100 \\
\Delta Q\% = 2\% \\
$
Hence, the heat energy radiated is increased by 2%

The correct option is D.

Note The rate at which the heat energy radiated by the hot body depends on area of the body and temperature. $\sigma $is the Stephen-Boltzmann constant whose value is \[5.670 \times {10^{ - 8}}W{m^{ - 2}}{K^{ - 4}}\]