Question

If $120\text{ J}$ of thermal energy is incident on area $3\text{ m}^2$, the amount of heat transmitted is $12\text{ J}$ , coefficient of absorption is $0.6$ , then the amount of heat reflected is

• A. $24\text{ J}$
• B. $30\text{ J}$
• C. $36\text{ J}$
• D. $40\text{ J}$

Hint:

For radiation: \[ a+r+t=1 \] Once you know any two coefficients, the third follows immediately.

Answer 1

The Correct Option is C

Concept:
If \(a\), \(r\), and \(t\) are absorptivity, reflectivity, and transmissivity, then: \[ a+r+t=1 \] ip

Step 1: Find transmissivity.
Incident heat: \[ Q_i=120\text{ J} \] Transmitted heat: \[ Q_t=12\text{ J} \] So, \[ t=\frac{12}{120}=0.1 \] ip

Step 2: Use absorptivity value.
Given: \[ a=0.6 \] Then, \[ r=1-a-t=1-0.6-0.1=0.3 \] ip

Step 3: Find reflected heat.
\[ Q_r = rQ_i = 0.3\times120 = 36\text{ J} \] ip Hence, the correct answer is:
\[ \boxed{(C)\ 36\text{ J}} \]