A cubical block of mass `1.0kg` and edge `5.0cm` is heated to `227^(@)C` . It is kept in an evacuated chamber maintained at `27^(@)C` . Assming that the block emits radiation like a blackbody, find the rate at which the temperature of the block will decreases. Specific heat capacity of the material of the block is `400Jkg^(-1)K^(-1)` .

Explanation

To solve the problem, we will follow these steps:

### Step 1: Calculate the surface area of the cubical block
The formula for the surface area \( A \) of a cube is given by:
\[ A = 6a^2 \]
where \( a \) is the edge length of the cube. Given that the edge length \( a = 5.0 \, \text{cm} = 0.05 \, \text{m} \), we can calculate the surface area:
\[ A = 6 \times (0.05)^2 = 6 \times 0.0025 = 0.015 \, \text{m}^2 \]

### Step 2: Convert the temperatures to Kelvin
The temperatures are given in Celsius, and we need to convert them to Kelvin:
- Temperature of the block \( T_1 = 227^\circ C = 227 + 273 = 500 \, K \)
- Surrounding temperature \( T_0 = 27^\circ C = 27 + 273 = 300 \, K \)

### Step 3: Use the Stefan-Boltzmann Law
According to the Stefan-Boltzmann Law, the rate of heat loss \( \frac{dQ}{dt} \) from a black body is given by:
\[ \frac{dQ}{dt} = \epsilon \sigma A (T_1^4 - T_0^4) \]
where:
- \( \epsilon = 1 \) (since it behaves like a black body)
- \( \sigma = 6.0 \times 10^{-8} \, \text{W/m}^2\text{K}^4 \) (Stefan-Boltzmann constant)
- \( A = 0.015 \, \text{m}^2 \)
- \( T_1 = 500 \, K \)
- \( T_0 = 300 \, K \)

### Step 4: Calculate the rate of heat loss
Now substituting the values into the equation:
\[ \frac{dQ}{dt} = 1 \times (6.0 \times 10^{-8}) \times 0.015 \times (500^4 - 300^4) \]

Calculating \( 500^4 \) and \( 300^4 \):
\[ 500^4 = 62500000000 \quad \text{and} \quad 300^4 = 8100000000 \]
Thus,
\[ 500^4 - 300^4 = 62500000000 - 8100000000 = 54400000000 \]

Now substituting this back:
\[ \frac{dQ}{dt} = (6.0 \times 10^{-8}) \times 0.015 \times 54400000000 \]
Calculating the above expression:
\[ \frac{dQ}{dt} = 6.0 \times 10^{-8} \times 0.015 \times 54400000000 = 4.905 \, \text{W} \]

### Step 5: Relate heat loss to temperature decrease
The heat lost can also be expressed in terms of the mass \( m \), specific heat capacity \( c \), and the change in temperature \( \frac{dT}{dt} \):
\[ \frac{dQ}{dt} = mc \frac{dT}{dt} \]
Where:
- \( m = 1 \, \text{kg} \)
- \( c = 400 \, \text{J/kg/K} \)

Setting the two expressions for \( \frac{dQ}{dt} \) equal to each other:
\[ 4.905 = 1 \times 400 \times \frac{dT}{dt} \]
Solving for \( \frac{dT}{dt} \):
\[ \frac{dT}{dt} = \frac{4.905}{400} = 0.0122625 \, \text{K/s} \approx 0.0123 \, \text{°C/s} \]

### Final Answer
The rate at which the temperature of the block will decrease is approximately:
\[ \frac{dT}{dt} \approx 0.0123 \, \text{°C/s} \]