Question

The temperature of the body is increased from $-{73}^{0}C$ to ${327}^{0}C$, the ratio of energy emitted per second is:

• A. 1:3
• B. 1:81
• C. 1:27
• D. 1:9

Answer 1

The correct option is B 1:81

We know that energy $\left(u\right)$ emmitted by a body is given by stefan law

i.e

$u=e\sigma A{T}^4$

here

$c\to$ emissivity

$\sigma \to$ stefans constnat

$A\to$ Surface area of body

$T\to$ temperature of body (in kalvin)

Now,

According to question

Initial temperature of body $=-{73}^{o}C$

$=200k$

$\Rightarrow u_1=e\sigma A(200{)}^4$

Final temperature $={327}^{o}C$

$=327+273=600K$

$u_2=c\sigma A(600{)}^4$

$\frac{u_1}{u_2}=\frac{e\sigma A(200{)}^4}{e\sigma A(600{)}^4}={\left(\frac{1}{3}\right)}^4$

$\Rightarrow \frac{u_1}{u_2}=\frac{1}{81}$

hence,

the ratio is $1:81$