Search any question & find its solution
Question:
Answered & Verified by Expert
If the temperature of a hot body is increased by $50 \%$, then the increase in the quantity of emitted heat radiation will be approximately
Options:
Solution:
1797 Upvotes
Verified Answer
The correct answer is:
$400 \%$
$\begin{aligned} & \text { Given: } \mathrm{T}_2=\mathrm{T}_1+\frac{50}{100} \mathrm{~T}_1 \\ & \mathrm{~T}_2=1.5 \mathrm{~T}_1\end{aligned}$
According to Stefan's law,
$$
\frac{\mathrm{Q}}{\mathrm{At}}=\sigma \mathrm{T}^4
$$
Percentage change in radiation is,
$$
\begin{aligned}
& \Delta \mathrm{E} \%=\frac{\mathrm{T}_2^4-\mathrm{T}_1^4}{\mathrm{~T}_1^4} \times 100 \\
& \Delta \mathrm{E} \%=\frac{(1.5)^4 \mathrm{~T}_1^4-\mathrm{T}_1^4}{\mathrm{~T}_1^4} \times 100 \\
& \Delta \mathrm{E} \% \approx 400 \%
\end{aligned}
$$
According to Stefan's law,
$$
\frac{\mathrm{Q}}{\mathrm{At}}=\sigma \mathrm{T}^4
$$
Percentage change in radiation is,
$$
\begin{aligned}
& \Delta \mathrm{E} \%=\frac{\mathrm{T}_2^4-\mathrm{T}_1^4}{\mathrm{~T}_1^4} \times 100 \\
& \Delta \mathrm{E} \%=\frac{(1.5)^4 \mathrm{~T}_1^4-\mathrm{T}_1^4}{\mathrm{~T}_1^4} \times 100 \\
& \Delta \mathrm{E} \% \approx 400 \%
\end{aligned}
$$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.