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 heat loss per unit area and unit time, we can use the concept of thermal conduction and the equation for calculating heat transfer.
The equation for heat transfer through conduction is:
Q = (k * A * ΔT) / L
Where:
Q = Heat transfer (in watts)
k = Thermal conductivity (in watts per meter-kelvin)
A = Area of the material (in square meters)
ΔT = Temperature difference (in kelvin or degrees Celsius)
L = Thickness of the material (in meters)
Given data:
- Ambient air temperature (Tamb) = 0°C (which is equivalent to 273K)
- Muscle temperature (Tmuscle) = 37°C (which is equivalent to 310K)
- Thermal conductivity of fat layer (kfat) = 0.53 W/mK
- Thickness of fat layer (Lfat) = 2 mm (which is equivalent to 0.002m)
- Thermal conductivity of epidermal layer (kepidermal) = 0.21 W/mK
- Thickness of epidermal layer (Lepidermal) = 0.1 mm (which is equivalent to 0.0001m)
Now, let's calculate the heat transfer through the fat layer and the epidermis layer separately, and then sum them up to get the total heat loss per unit area and unit time.
1. Heat transfer through the fat layer:
ΔTfat = Tmuscle - Tamb = 310K - 273K = 37K
Qfat = (kfat * A * ΔTfat) / Lfat
= (0.53 * A * 37) / 0.002
= 19.85 * A
2. Heat transfer through the epidermal layer:
ΔTepidermal = Tmuscle - Tamb = 310K - 273K = 37K
Qepidermal = (kepidermal * A * ΔTepidermal) / Lepidermal
= (0.21 * A * 37) / 0.0001
= 777 * A
3. Total heat loss per unit area and unit time:
Qtotal = Qfat + Qepidermal
= 19.85 * A + 777 * A
= (19.85 + 777) * A
= 796.85 * A
Therefore, the heat loss per unit area and unit time from the muscle to the ambient air is 796.85 times the area (A).