Assume the muscle is 37C and is separated from the outside air by layers of fat and skin. The layer of fat, at a particular location on the skin, is 2 mm thick and has a conductivity of 0.53 W/mK. Finally, the outermost epidermal layer is 0.1 mm thick with a thermal conductivity of 0.21 W/mK. How much heat is lost per unit area and unit time if the ambient air temperature is 0C?
To calculate the amount of heat lost per unit area and unit time, we need to consider the heat conduction through the layers of fat and skin.
Let's assume the temperature difference between the muscle (37°C) and the ambient air (0°C) is ΔT = (37 - 0) = 37°C.
First, we need to calculate the overall thermal resistance for the layers of fat and skin. The thermal resistance (R) is given by the equation:
R = thickness / (thermal conductivity * area)
For the layer of fat:
R_fat = (2 mm) / (0.53 W/mK * 0.001 m/mm * 1 m^2) = 3.774 m^2K/W
For the outermost epidermal layer:
R_epidermis = (0.1 mm) / (0.21 W/mK * 0.001 m/mm * 1 m^2) = 0.476 m^2K/W
Next, we can calculate the total thermal resistance for the layers:
R_total = R_fat + R_epidermis = 3.774 m^2K/W + 0.476 m^2K/W = 4.25 m^2K/W
The rate of heat transfer (Q) per unit area (A) and unit time (t) is given by the equation:
Q = (ΔT / R_total) * A
Here, A represents the surface area of the skin.
Let's assume a standard value for the surface area of the skin, such as 1 m^2.
Substituting the values:
Q = (37°C / 4.25 m^2K/W) * 1 m^2 = 8.706 W
Therefore, the heat lost per unit area and unit time is approximately 8.706 Watts.