6 Questions around this concept.
Q amount of heat falls on a surface. 20% heat is absorbed and 35% is reflected, then the transmittance is
Radiation - The process of the transfer of heat from one place to another place without any requirement of the medium is called radiation. It means that the radiation does not need any material medium to propagate.
Characteristics of Radiation -
When thermal radiations (Q) fall on a body, they are partly reflected, partly absorbed, and partly transmitted as shown in the below figure.

So we can write
$$
\begin{aligned}
& Q=Q_a+Q_t+Q_r \\
& \quad \frac{Q}{Q}=\frac{Q_a}{Q}+\frac{Q_t}{Q}+\frac{Q_r}{r}
\end{aligned}
$$
or $\quad 1=a+r+t$
Where
$$
\begin{aligned}
& \frac{Q_a}{Q}=a=\text { Absorptance } \\
& \frac{Q_r}{Q}=r=\text { Reflactance } \\
& \frac{Q_t}{Q}=t=\text { Transmittance }
\end{aligned}
$$
So
If r = t = 0 and a = 1 then a body is a perfectly black body.
If, a = r = 0 and t = 1 the body is perfect transmitter
If $t=0 \Rightarrow r+a=1$ or $a=1-r$
i.e. good reflectors are bad absorbers.
I. If surrounding temperature= body temperature
then $Q_{\text {emmition }}=Q_{\text {absorbed }}$
i.e the body will emit and absorb at the same rate
the temperature of the body remains constant (thermal equilibrium)
II. If surrounding temperature > body temperature
then $Q_{\text {emmition }}<Q_{\text {absorbed }}$
i.e. temperature of the body increases and it appears hotter.
III. If surrounding temperature < body temperature
then $Q_{\text {emmition }}>Q_{\text {absorbed }}$
i.e. temperature of the body decreases and consequently the body appears colder.
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