Seals may cool themselves by using thermal windows, patches on their bodies with much higher than average surface temperature. Suppose a seal has a 0.030 m^2 thermal window at a temperature of 30 degree C. If the seal’s surroundings are a frosty -10 degree C, what is the net rate of energy loss by radiation? Assume an emissivity equal to that of a human.
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To calculate the net rate of energy loss by radiation, we will use the Stefan-Boltzmann Law:
Net energy loss = εσA(T₁⁴ - T₂⁴)
Where:
ε is the emissivity (equal to 1 for a perfect black body and assumed to be the same as a human)
σ is the Stefan-Boltzmann constant (5.67 x 10^-8 W/(m²K⁴))
A is the area of the thermal window (0.030 m²)
T₁ is the temperature of the thermal window (30 °C + 273.15 = 303.15 K)
T₂ is the temperature of the surroundings (-10 °C + 273.15 = 263.15 K)
Now, let's plug in the values and calculate the net rate of energy loss:
Net energy loss = 1 × (5.67 x 10^-8) × 0.030 × (303.15⁴ - 263.15⁴)
Note: We converted the temperatures to Kelvin because the Stefan-Boltzmann constant is expressed in terms of Kelvin.
Calculating this expression will give us the net rate of energy loss by radiation for the seal.