A black rectangular surface of area A emits energy E per second at `27^circC`. If length and breadth are reduced to one third of initial value and temperature is raised to `327^circC`, then energy emitted per second becomes

Options

A. `(4E)/(9)`
B. `(7E)/(9)`
C. `(10E)/(9)`
D. `(16E)/(9)`

Answer

To solve the problem, we need to determine the new energy emitted per second by a black rectangular surface after its dimensions and temperature have changed.

### Step 1: Understand the initial conditions
The initial area of the surface is \( A \) and it emits energy \( E \) per second at a temperature of \( 27^\circ C \).

### Step 2: Convert the initial temperature to Kelvin
The initial temperature in Kelvin is:
\[ T = 27 + 273 = 300 \, K \]

### Step 3: Determine the new dimensions and area
The length and breadth of the surface are reduced to one third of their initial values. Thus, the new dimensions will be:
\[ \text{New Length} = \frac{L}{3}, \quad \text{New Breadth} = \frac{B}{3} \]
The new area \( A' \) can be calculated as:
\[ A' = \left(\frac{L}{3}\right) \times \left(\frac{B}{3}\right) = \frac{LB}{9} = \frac{A}{9} \]

### Step 4: Convert the new temperature to Kelvin
The new temperature is raised to \( 327^\circ C \):
\[ T' = 327 + 273 = 600 \, K \]

### Step 5: Use the Stefan-Boltzmann Law
According to the Stefan-Boltzmann Law, the energy emitted per second \( E \) is given by:
\[ E = \epsilon \sigma A T^4 \]
where \( \epsilon \) is the emissivity (which remains constant), \( \sigma \) is the Stefan-Boltzmann constant, \( A \) is the area, and \( T \) is the absolute temperature.

### Step 6: Write the expressions for initial and new energy emitted
For the initial conditions:
\[ E = \epsilon \sigma A (300)^4 \]
For the new conditions:
\[ E' = \epsilon \sigma A' (600)^4 \]

### Step 7: Substitute the new area and temperature
Substituting \( A' = \frac{A}{9} \) and \( T' = 600 \, K \):
\[ E' = \epsilon \sigma \left(\frac{A}{9}\right) (600)^4 \]

### Step 8: Relate \( E' \) to \( E \)
Now, we can express \( E' \) in terms of \( E \):
\[ E' = \epsilon \sigma \left(\frac{A}{9}\right) (600)^4 = \frac{1}{9} \epsilon \sigma A (600)^4 \]
We know \( E = \epsilon \sigma A (300)^4 \), so we can substitute:
\[ E' = \frac{1}{9} E \left(\frac{(600)^4}{(300)^4}\right) \]

### Step 9: Simplify the temperature ratio
Calculating the temperature ratio:
\[ \frac{(600)^4}{(300)^4} = \left(\frac{600}{300}\right)^4 = 2^4 = 16 \]

### Step 10: Final expression for \( E' \)
Now substituting back:
\[ E' = \frac{1}{9} E \times 16 = \frac{16}{9} E \]

### Conclusion
Thus, the energy emitted per second after the changes is:
\[ E' = \frac{16}{9} E \]