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

Step 1: Understanding the Concept:
Incident heat energy is conservation: $Q_{incident} = Q_{absorbed} + Q_{reflected} + Q_{transmitted}$.
Step 2: Key Formula or Approach:
$a + r + t = 1$, where $a = Q_a/Q_i$, $r = Q_r/Q_i$, $t = Q_t/Q_i$.
Step 3: Detailed Explanation:
$Q_a = 0.6 \times 120 = 72\text{ J}$.
$Q_t = 12\text{ J}$.
$Q_r = 120 - 72 - 12 = 36\text{ J}$.
Step 4: Final Answer:
Amount of heat reflected is $36\text{ J}$.