An incandescent lamp consumint `P=54W` is immersed into a transparent calorimeter containing `V=10^3cm^3` of water in 420 s the water is heated by `4^@C`. The percentage of the energy consumed by the lamp that passes out of the calorimeter in the form of radiant energy is

Options

A. `81.5%`
B. `26%`
C. `40.5%`
D. `51.5%`

Explanation

To solve the problem step by step, we will calculate the percentage of energy consumed by the incandescent lamp that passes out of the calorimeter as radiant energy.

### Step 1: Identify the given values
- Power of the lamp, \( P = 54 \, \text{W} \)
- Volume of water, \( V = 10^3 \, \text{cm}^3 = 10^{-3} \, \text{m}^3 \)
- Time, \( t = 420 \, \text{s} \)
- Temperature rise, \( \Delta T = 4 \, ^\circ C \)

### Step 2: Calculate the mass of the water
The mass of the water can be calculated using the formula:
\[ m = V \times \text{density} \]
The density of water is approximately \( 1000 \, \text{kg/m}^3 \). Thus,
\[ m = 10^{-3} \, \text{m}^3 \times 1000 \, \text{kg/m}^3 = 1 \, \text{kg} \]

### Step 3: Calculate the heat absorbed by the water
The heat absorbed by the water can be calculated using the formula:
\[ Q = m \times s \times \Delta T \]
where \( s \) is the specific heat capacity of water, approximately \( 4200 \, \text{J/(kg} \cdot ^\circ C) \). Therefore,
\[ Q = 1 \, \text{kg} \times 4200 \, \text{J/(kg} \cdot ^\circ C) \times 4 \, ^\circ C = 16800 \, \text{J} \]

### Step 4: Calculate the power consumed by the water
The power consumed by the water can be calculated as:
\[ P_{\text{water}} = \frac{Q}{t} \]
Substituting the values we found:
\[ P_{\text{water}} = \frac{16800 \, \text{J}}{420 \, \text{s}} = 40 \, \text{W} \]

### Step 5: Calculate the power not absorbed by the water
The power not absorbed by the water (which is the power that goes out as radiant energy) can be calculated as:
\[ P_{\text{radiant}} = P - P_{\text{water}} = 54 \, \text{W} - 40 \, \text{W} = 14 \, \text{W} \]

### Step 6: Calculate the percentage of energy consumed by the lamp that passes out as radiant energy
The percentage of energy that passes out as radiant energy can be calculated using the formula:
\[ \text{Percentage} = \left( \frac{P_{\text{radiant}}}{P} \right) \times 100 \]
Substituting the values:
\[ \text{Percentage} = \left( \frac{14 \, \text{W}}{54 \, \text{W}} \right) \times 100 \approx 25.93\% \]
Rounding off, we can say approximately \( 26\% \).

### Final Answer
The percentage of the energy consumed by the lamp that passes out of the calorimeter in the form of radiant energy is approximately \( 26\% \).

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