Question 682: Thermodynamics & KTG

Question

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

Accepted Answer

We are given:
Thermal energy incident = $Q_	ext{incident} = 120 \, 	ext{J}$
Heat transmitted = $Q_	ext{transmitted} = 12 \, 	ext{J}$
Coefficient of absorption = $a = 0.6$
Step 1: Absorbed heat
Coefficient of absorption is
$ a = \frac{Q_	ext{absorbed}}{Q_	ext{incident}} $
So,
$ Q_	ext{absorbed} = a \cdot Q_	ext{incident} = 0.6 	imes 120 = 72 \, 	ext{J} $
Step 2: Energy conservation
Total incident energy is partitioned as:
$ Q_	ext{incident} = Q_	ext{absorbed} + Q_	ext{reflected} + Q_	ext{transmitted} $
So,
$ 120 = 72 + Q_	ext{reflected} + 12 $
$ Q_	ext{reflected} = 120 - 84 = 36 \, 	ext{J} $
Ã¢ÂÂ Answer:
The amount of heat reflected is
36 J
Correct Option: C (36 J)

Answer

24 J

Answer

30 J

Answer

36 J

Answer

40 J

Thermodynamics & KTG

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

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