To solve the problem, we will follow these steps:
### Step 1: Convert the mass and area to appropriate units
- The mass of the black body is given as \(34.38 \, \text{g}\). We convert this to kilograms:
\[
m = 34.38 \, \text{g} = 0.03438 \, \text{kg}
\]
- The surface area is given as \(19.2 \, \text{cm}^2\). We convert this to square meters:
\[
A = 19.2 \, \text{cm}^2 = 19.2 \times 10^{-4} \, \text{m}^2 = 0.00192 \, \text{m}^2
\]
### Step 2: Identify the given values
- Initial temperature \(T = 400 \, \text{K}\)
- Constant temperature of the enclosure \(T_0 = 300 \, \text{K}\)
- Rate of cooling \(dT/dt = 0.04 \, \text{°C/s} = 0.04 \, \text{K/s}\)
- Stefan's constant \(\sigma = 5.73 \times 10^{-8} \, \text{W/m}^2\text{K}^4\)
### Step 3: Use the formula for rate of cooling
The rate of cooling can be expressed using Newton's law of cooling:
\[
\frac{dT}{dt} = -\frac{E \cdot A \cdot \sigma (T^4 - T_0^4)}{m \cdot C}
\]
Where:
- \(E\) is the emissivity (for a black body, \(E = 1\))
- \(C\) is the specific heat capacity we want to find.
### Step 4: Rearranging the formula
Rearranging the formula to solve for \(C\):
\[
C = -\frac{E \cdot A \cdot \sigma (T^4 - T_0^4)}{m \cdot \frac{dT}{dt}}
\]
### Step 5: Substitute the known values
Substituting the known values into the equation:
\[
C = -\frac{1 \cdot (0.00192) \cdot (5.73 \times 10^{-8}) \cdot (400^4 - 300^4)}{0.03438 \cdot (-0.04)}
\]
### Step 6: Calculate \(T^4 - T_0^4\)
Calculating \(400^4\) and \(300^4\):
\[
400^4 = 256 \times 10^8 \quad \text{and} \quad 300^4 = 81 \times 10^8
\]
So,
\[
400^4 - 300^4 = (256 - 81) \times 10^8 = 175 \times 10^8
\]
### Step 7: Plugging in the values
Now substituting back:
\[
C = \frac{(0.00192) \cdot (5.73 \times 10^{-8}) \cdot (175 \times 10^8)}{0.03438 \cdot 0.04}
\]
### Step 8: Calculate \(C\)
Calculating the numerator:
\[
0.00192 \cdot 5.73 \times 10^{-8} \cdot 175 \times 10^8 = 0.00192 \cdot 5.73 \cdot 175 = 0.00192 \cdot 1003.75 = 1.926 \, \text{J}
\]
Calculating the denominator:
\[
0.03438 \cdot 0.04 = 0.0013752
\]
Now substituting these values:
\[
C = \frac{1.926}{0.0013752} \approx 1400 \, \text{J/kg/K}
\]
### Final Answer
Thus, the specific heat \(C\) of the body is approximately \(1400 \, \text{J/kg/K}\).
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