If 150 J of energy is incident on area ` 2m^(2)` . If `Q_(r)=15J`, coefficient of absorption is 0.6 , then amount of energy transmitted is

Options

A. `50J`
B. `45 J`
C. `40 J`
D. `30 J`

Answer

To solve the problem step by step, we will use the given data and apply the principles of energy absorption, transmission, and radiation.

### Step-by-Step Solution:

1. **Identify the Given Data:**
- Total energy incident (Q) = 150 J
- Area = 2 m² (not directly needed for the calculation)
- Radiated heat (Q_r) = 15 J
- Coefficient of absorption (α) = 0.6

2. **Understanding the Energy Distribution:**
The total energy incident on the surface can be divided into three parts:
- Radiated energy (Q_r)
- Absorbed energy (Q_a)
- Transmitted energy (Q_t)

The relationship can be expressed as:
\[ Q = Q_r + Q_a + Q_t \]

3. **Expressing the Coefficients:**
We can express the coefficients in terms of the total energy:
- Coefficient of radiated energy: \(\frac{Q_r}{Q}\)
- Coefficient of absorbed energy: \(\frac{Q_a}{Q}\)
- Coefficient of transmitted energy: \(\frac{Q_t}{Q}\)

From the problem, we know:
\[ \frac{Q_r}{Q} + \frac{Q_a}{Q} + \frac{Q_t}{Q} = 1 \]

4. **Substituting Known Values:**
We already have:
- \(Q_r = 15 \, \text{J}\)
- \(Q = 150 \, \text{J}\)
- Coefficient of absorption (α) = 0.6

Now substituting \(Q_r\) and α into the equation:
\[ \frac{15}{150} + 0.6 + \frac{Q_t}{150} = 1 \]

5. **Calculating the Coefficient of Radiated Energy:**
\[ \frac{15}{150} = 0.1 \]

6. **Setting Up the Equation:**
Substitute the value back into the equation:
\[ 0.1 + 0.6 + \frac{Q_t}{150} = 1 \]

7. **Simplifying the Equation:**
Combine the coefficients:
\[ 0.7 + \frac{Q_t}{150} = 1 \]

8. **Isolating \(Q_t\):**
Subtract 0.7 from both sides:
\[ \frac{Q_t}{150} = 1 - 0.7 \]
\[ \frac{Q_t}{150} = 0.3 \]

9. **Calculating \(Q_t\):**
Multiply both sides by 150:
\[ Q_t = 0.3 \times 150 = 45 \, \text{J} \]

### Final Answer:
The amount of energy transmitted (Q_t) is **45 J**.