Stanley with skin temperature of 37°c rest in a room where the temperature of the wall is 30°c. Determine the net rate of heat loss as a result of radiation if the Emissivity is 1 while the surface area of the body is 1.6^2m?
To determine the net rate of heat loss due to radiation, we can use the Stefan-Boltzmann Law. The equation for the rate of heat loss due to radiation is given by:
Q = ε * σ * A * (T₁⁴ - T₂⁴)
Where:
Q = Net rate of heat loss
ε = Emissivity (which is 1 in this case)
σ = Stefan-Boltzmann constant (approximately 5.67 × 10⁻⁸ W/m²K⁴)
A = Surface area of the body (1.6² = 2.56 m²)
T₁ = Temperature of the body (37°C converted to Kelvin = 37 + 273 = 310 K)
T₂ = Temperature of the surroundings (30°C converted to Kelvin = 30 + 273 = 303 K)
Plugging in the values into the equation:
Q = 1 * (5.67 × 10⁻⁸) * 2.56 * (310⁴ - 303⁴)
Calculating this gives us:
Q ≈ 1 * (5.67 × 10⁻⁸) * 2.56 * (9238000 - 8478801)
Simplifying further:
Q ≈ 1 * (5.67 × 10⁻⁸) * 2.56 * 759199
Finally, evaluating this expression yields the net rate of heat loss due to radiation:
Q ≈ 0.0000953 W or 9.53 × 10⁻⁵ W
Therefore, the net rate of heat loss as a result of radiation is approximately 9.53 × 10⁻⁵ W.