Question 692: Thermodynamics & KTG

Question

A rectangular block of surface area A emits energy E per second at $27^{\circ} \mathrm{C}$. If length and breadth is reduced to half of initial value and temperature is raised to $327^{\circ} \mathrm{C}$ then energy emitted per second becomes

Accepted Answer

Let us solve this step-by-step using
Stefan-Boltzmann Law
, which is discussed in your NCERT textbook.
Step 1: Stefan-Boltzmann Law
The power radiated by a body (energy emitted per second) is given by:
$$ P = \sigma \epsilon A T^4 $$
where:
$\sigma$ = Stefan-Boltzmann constant,
$\epsilon$ = emissivity,
$A$ = surface area,
$T$ = absolute temperature in Kelvin.
Step 2: Initial Condition
Let the initial emission be $E$ at $27^\circ \mathrm{C}$ (which is $300\,\mathrm{K}$):
$$ E = \sigma \epsilon A (300)^4 $$
Step 3: New Condition
Length and breadth are reduced to half, so area $A' = \left(\frac{1}{2}ight) 	imes \left(\frac{1}{2}ight) A = \frac{1}{4}A$
Temperature is raised to $327^\circ \mathrm{C} = 600\,\mathrm{K}$
The new emission is:
$$ E' = \sigma \epsilon A' (600)^4 = \sigma \epsilon \left(\frac{A}{4}ight) (600)^4 $$
Step 4: Find the Ratio
Take the ratio $\dfrac{E'}{E}$:
$$ \frac{E'}{E} = \frac{\sigma \epsilon \frac{A}{4} (600)^4}{\sigma \epsilon A (300)^4} = \frac{1}{4} \frac{(600)^4}{(300)^4} $$
Since $600 = 2 	imes 300$,
$$ (600)^4 = (2 	imes 300)^4 = 2^4 	imes (300)^4 = 16 	imes (300)^4 $$
Therefore,
$$ \frac{E'}{E} = \frac{1}{4} 	imes \frac{16 	imes (300)^4}{(300)^4} = \frac{1}{4} 	imes 16 = 4 $$
$$ E' = 4E $$
Final Answer
Option B: $4E$

Answer

2 E

Answer

4 E

Answer

E

Answer

8 E

Thermodynamics & KTG

A rectangular block of surface area A emits energy E per second at 27^{\circ} \mathrm{C}. If length and breadth is reduced to half of initial value and temperature is raised to 327^{\circ} \mathrm{C} then energy emitted per second becomes

Choose the correct answer: