The thickness of a metallic plate is 0.4 cm. The temperature between its two surfaces is `20^(@)C`. The quantity of heat flowing per second is 50 calories from `5cm^(2)` area. In CGS system, the coefficient of thermal conductivity will be

Options

A. 0.4
B. 0.6
C. 0.2
D. 0.5

Explanation

To find the coefficient of thermal conductivity (K) of the metallic plate, we can use the formula for the rate of heat transfer through a material:

\[ \frac{Q}{t} = \frac{K \cdot A \cdot (T_1 - T_2)}{L} \]

Where:
- \( \frac{Q}{t} \) is the rate of heat transfer (in calories per second),
- \( K \) is the coefficient of thermal conductivity (in CGS units, calories per second per degree Celsius per centimeter),
- \( A \) is the area through which heat is being transferred (in cm²),
- \( (T_1 - T_2) \) is the temperature difference (in degrees Celsius),
- \( L \) is the thickness of the material (in cm).

### Step 1: Identify the given values
- Thickness of the plate, \( L = 0.4 \) cm
- Temperature difference, \( T_1 - T_2 = 20 \) °C
- Area, \( A = 5 \) cm²
- Rate of heat transfer, \( \frac{Q}{t} = 50 \) calories/second

### Step 2: Substitute the values into the formula
We can rearrange the formula to solve for \( K \):

\[ K = \frac{Q/t \cdot L}{A \cdot (T_1 - T_2)} \]

Substituting the known values:

\[ K = \frac{50 \cdot 0.4}{5 \cdot 20} \]

### Step 3: Calculate the numerator and denominator
Calculating the numerator:

\[ 50 \cdot 0.4 = 20 \]

Calculating the denominator:

\[ 5 \cdot 20 = 100 \]

### Step 4: Divide the numerator by the denominator
Now, we can find \( K \):

\[ K = \frac{20}{100} = 0.2 \]

### Step 5: State the units
In CGS units, the coefficient of thermal conductivity \( K \) is expressed as:

\[ K = 0.2 \text{ calories per second per degree Celsius per centimeter} \]

### Final Answer
Thus, the coefficient of thermal conductivity is:

\[ K = 0.2 \, \text{calories/(s°C cm)} \]

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