A metal plate 4 mm thick has a temperature difference of `32^@C` between its faces. It transmits 200 kcal`//`h through an area of 5 `cm^2` Thermal conductivity of the material is

Options

A. `58.33 W//m-^@C`
B. 33.58 `W//m-^@C`
C. `5xx 10^(-4) W//m-^@C`
D. None of these

Explanation

To find the thermal conductivity (K) of the metal plate, we can use the formula for heat transfer through conduction, which is given by:

\[ \frac{Q}{T} = K \cdot A \cdot \frac{\Delta T}{\Delta L} \]

Where:
- \(Q/T\) is the heat transfer rate (in kcal/h),
- \(K\) is the thermal conductivity (in W/m·°C),
- \(A\) is the area (in m²),
- \(\Delta T\) is the temperature difference (in °C),
- \(\Delta L\) is the thickness of the plate (in m).

### Step-by-Step Solution:

1. **Convert the given values to appropriate units:**
- Thickness (\(\Delta L\)): 4 mm = \(4 \times 10^{-3}\) m
- Temperature difference (\(\Delta T\)): 32 °C (no conversion needed)
- Area (\(A\)): 5 cm² = \(5 \times 10^{-4}\) m²
- Heat transfer rate (\(Q/T\)): 200 kcal/h. We need to convert this to watts (W).
- 1 kcal = 4184 J
- 1 hour = 3600 seconds
- Therefore, \(200 \text{ kcal/h} = 200 \times 4184 \text{ J/h} = 200 \times 4184 / 3600 \text{ W}\)

2. **Calculate the heat transfer rate in watts:**
\[ Q/T = \frac{200 \times 4184}{3600} = \frac{836800}{3600} \approx 232.22 \text{ W} \]

3. **Substitute the values into the heat transfer equation to find \(K\):**
\[ K = \frac{Q/T \cdot \Delta L}{A \cdot \Delta T} \]
Substituting the values:
\[ K = \frac{232.22 \cdot (4 \times 10^{-3})}{(5 \times 10^{-4}) \cdot 32} \]

4. **Calculate \(K\):**
\[ K = \frac{232.22 \cdot 0.004}{0.0005 \cdot 32} \]
\[ K = \frac{0.92888}{0.016} = 58.68 \text{ W/m·°C} \]

5. **Final Result:**
The thermal conductivity \(K\) of the material is approximately \(58.68 \text{ W/m·°C}\).