(a) Calculate the rate at which body heat is conducted through the clothing of a skier in a steady-state process, given the following data`:` the body surface area is `1.8m^(2)`, and the clothing is 1.0 cm thick , the skin surface temperature is `33^(@)C` and the outer surface of the clothing is at `1.0^(@)C`, the thermal conductivity of the clothing is `0.040 W//m . K.` (b) If , after a fall, the skier's clothes became soaked with water of thermal conductivity `0.60 W//m.K`, by how much is the rate of conduction multiplied ?

Answer

To solve the problem step by step, we will use the formula for heat conduction through a material, which is given by Fourier's law of heat conduction:

\[ Q = \frac{k \cdot A \cdot (T_1 - T_2)}{d} \]

where:
- \( Q \) is the rate of heat transfer (in watts, W),
- \( k \) is the thermal conductivity of the material (in W/m·K),
- \( A \) is the surface area (in m²),
- \( T_1 \) and \( T_2 \) are the temperatures on either side of the material (in °C),
- \( d \) is the thickness of the material (in m).

### Part (a)

1. **Identify the given values:**
- Body surface area, \( A = 1.8 \, m^2 \)
- Thickness of clothing, \( d = 1.0 \, cm = 0.01 \, m \)
- Skin surface temperature, \( T_1 = 33 \, °C \)
- Outer surface temperature, \( T_2 = 1 \, °C \)
- Thermal conductivity of clothing, \( k = 0.040 \, W/m·K \)

2. **Calculate the temperature difference:**
\[ \Delta T = T_1 - T_2 = 33 - 1 = 32 \, °C \]

3. **Substitute the values into the heat conduction formula:**
\[ Q = \frac{0.040 \cdot 1.8 \cdot 32}{0.01} \]

4. **Calculate the numerator:**
\[ 0.040 \cdot 1.8 \cdot 32 = 2.304 \, W \]

5. **Now divide by the thickness:**
\[ Q = \frac{2.304}{0.01} = 230.4 \, W \]

Thus, the rate at which body heat is conducted through the clothing of the skier is **230.4 W**.

### Part (b)

1. **Identify the new thermal conductivity of water:**
- \( k_{water} = 0.60 \, W/m·K \)

2. **Use the same formula for heat conduction with the new thermal conductivity:**
\[ Q_{water} = \frac{0.60 \cdot 1.8 \cdot 32}{0.01} \]

3. **Calculate the numerator:**
\[ 0.60 \cdot 1.8 \cdot 32 = 34.56 \, W \]

4. **Now divide by the thickness:**
\[ Q_{water} = \frac{34.56}{0.01} = 3456 \, W \]

5. **Calculate the multiplication factor:**
\[ \text{Multiplication factor} = \frac{Q_{water}}{Q} = \frac{3456}{230.4} = 15.0 \]

Thus, the rate of conduction is multiplied by **15** when the skier's clothes become soaked with water.