The total energy radiated from a block body source at constant temperature is collected for one minute and is used to heat a quantity of water. The temperature of water is found to increase from `20^(@)C` to `20.5^(@)C`. If the absolute temperature of the blackbody is doubled and the experiment is repeated with the same quantity of water of `20^(@)C`, the temperature of water will be:

Options

A. `21^(@)C`
B. `22^(@)C`
C. `24^(@)C`
D. `28^(@)C`

Answer

To solve the problem, let's break it down step by step.

### Step 1: Understand the initial situation
We have a blackbody radiating energy at a certain absolute temperature \( T \). The energy collected from this blackbody for one minute raises the temperature of a certain quantity of water from \( 20^\circ C \) to \( 20.5^\circ C \).

### Step 2: Use the Stefan-Boltzmann Law
According to the Stefan-Boltzmann Law, the power radiated by a blackbody is given by:
\[ P = \sigma A T^4 \]
where \( \sigma \) is the Stefan-Boltzmann constant, \( A \) is the area of the blackbody, and \( T \) is the absolute temperature in Kelvin.

### Step 3: Relate energy to temperature change in water
The energy supplied to the water can be expressed as:
\[ Q = mc\Delta T \]
where \( m \) is the mass of the water, \( c \) is the specific heat capacity of water, and \( \Delta T \) is the change in temperature of the water.

In the first case, the change in temperature \( \Delta T \) is:
\[ \Delta T = 20.5 - 20 = 0.5^\circ C \]

### Step 4: Set up the equation for the first scenario
The energy radiated by the blackbody in one minute is equal to the heat absorbed by the water:
\[ \sigma A T^4 \cdot t = mc(0.5) \]
where \( t \) is the time (1 minute).

### Step 5: Consider the second scenario with doubled temperature
Now, if the absolute temperature of the blackbody is doubled, the new temperature is \( 2T \). The power radiated now becomes:
\[ P' = \sigma A (2T)^4 = \sigma A \cdot 16 T^4 \]

### Step 6: Set up the equation for the second scenario
The energy radiated in one minute at the new temperature is:
\[ \sigma A (2T)^4 \cdot t = mc(\Delta T') \]
where \( \Delta T' \) is the new change in temperature of the water.

### Step 7: Relate the two scenarios
From the first scenario, we have:
\[ \sigma A T^4 \cdot t = mc(0.5) \]
From the second scenario:
\[ \sigma A \cdot 16 T^4 \cdot t = mc(\Delta T') \]

### Step 8: Divide the second equation by the first
\[ \frac{\sigma A \cdot 16 T^4 \cdot t}{\sigma A T^4 \cdot t} = \frac{mc(\Delta T')}{mc(0.5)} \]
This simplifies to:
\[ 16 = \frac{\Delta T'}{0.5} \]

### Step 9: Solve for \( \Delta T' \)
\[ \Delta T' = 16 \times 0.5 = 8^\circ C \]

### Step 10: Find the final temperature of the water
The initial temperature of the water is \( 20^\circ C \), so the final temperature will be:
\[ T_f = 20 + 8 = 28^\circ C \]

### Final Answer
The final temperature of the water when the absolute temperature of the blackbody is doubled is \( 28^\circ C \).

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