Suppose the skin temperature of a naked person is 34 degree celsius when the person is standing inside a room whose temperature is 23 celsius. The skin area of the individual is 1.5m^2.
(a) Assuming the emissivity is 0.80, find the net loss of radint power from the body.
(b) Determine the number of food Calories of energy(1 food calorie=4186J) that is lost in one hour due to the net loss rate obtained in part (a0. Metabolic conversion of food into energy replaces this loss.
I will be happy to critique your thinking.
Ignore convection losses, which in fact are probably greater than radiation.
Yeahh Go UBC
To find the net loss of radiant power from the body, we can use the Stefan-Boltzmann law. The Stefan-Boltzmann law states that the power radiated by a black body is proportional to the fourth power of its temperature. The formula is as follows:
P = εσA(T₁^4 - T₂^4)
where:
P is the power radiated (in watts),
ε is the emissivity (a dimensionless quantity between 0 and 1),
σ is the Stefan-Boltzmann constant (approximately equal to 5.67 x 10^-8 W/m^2K^4),
A is the surface area of the body (in square meters),
T₁ is the temperature of the body (in Kelvin),
T₂ is the temperature of the surrounding (in Kelvin).
First, let's convert the temperatures from degree Celsius to Kelvin:
Skin temperature (T₁) = 34°C + 273.15 = 307.15 K
Room temperature (T₂) = 23°C + 273.15 = 296.15 K
Now, we can calculate the net loss of radiant power using the given values:
ε = 0.80 (emissivity)
A = 1.5 m² (skin area)
P = 0.80 * 5.67 x 10^-8 * 1.5 * (307.15^4 - 296.15^4)
Note: Make sure to use Kelvin for the temperatures in the calculation.
To determine the number of food Calories (kcal) of energy lost in one hour, we need to convert the power in watts to energy in joules, and then to food Calories (kcal). The conversion factor is 1 food calorie = 4186 joules.
Energy = Power * Time
Energy (in joules) = P * 3600 seconds (1 hour = 3600 seconds)
Finally, we can convert the energy from joules to food Calories (kcal):
Energy (in kcal) = Energy (in joules) / 4186
That's it! Plug in the values into the formulas to find the net loss of radiant power and the energy loss in food Calories.