Suppose the skin temperature of a naked person is 34°C when the person is standing inside a room whose temperature is 25°C. The skin area of the individual is 2.0 m2

a) Assuming the emissivity is 0.80, find the net loss of radiant power from the body

b) Determine the number of food Calories of energy (1 food Calorie = 4186 J) that is lost in one hour due to the net loss rate obtained in part (a). Metabolic conversion of food into energy replaces this loss.

I used
a)
Q/t = emissivity x stefan-boltzmann constant x T^4 x A

= 0.8 X 5.67^-8 X 298.15^4 X 2

=719.277

b) 1 watt per hour = 3600J
total joules = 3600 X 719.277 = 2589399.087

2589399.087 /4186 = total calories

Answers are wrong

A food calorie is 1000energy calories.

ok. what about the (a) part?

a) Q/t = emissivity x stefan-boltzmann constant x (T1^4 - T2^4) x A

= 0.8 X 5.67^-8 X (310.15^4 - 298.15^4) X 2

=719.277 W

Well, it seems like you're feeling "heated" about the wrong answer. But don't worry, I'm here to help you chill out! Let's take another crack at it, shall we?

a) To find the net loss of radiant power, we can use the formula:

Q/t = ε x σ x (T^4 - T0^4) x A

Where:
Q/t is the net loss of radiant power
ε is the emissivity (0.80 in this case)
σ is the Stefan-Boltzmann constant (5.67 x 10^-8 W/m^2K^4)
T is the skin temperature (34°C + 273.15 = 307.15 K)
T0 is the room temperature (25°C + 273.15 = 298.15 K)
A is the skin area (2.0 m^2)

Now, let's plug in the numbers:

Q/t = 0.80 x (5.67 x 10^-8 W/m^2K^4) x (307.15^4 - 298.15^4) x 2.0

After some calculations, you should get the net loss of radiant power in watts!

b) To determine the energy lost in one hour, we can simply multiply the net loss rate obtained in part (a) by 3600 (since there are 3600 seconds in an hour):

Energy lost = Net loss rate x 3600 J

Now, let's calculate the number of "food Calories" of energy lost. Remember, 1 food Calorie is equal to 4186 J:

Food Calories lost = (Net loss rate x 3600 J) / 4186

With these calculations, we'll have a better understanding of the energy losses. Just make sure to keep your cool!

To calculate the net loss of radiant power from the body, you can use the Stefan-Boltzmann Law:

Q/t = εσA(T² - T₀²)

Where:
Q/t is the net loss of radiant power (in Watts)
ε is the emissivity (given as 0.80)
σ is the Stefan-Boltzmann constant (5.67 x 10^-8 W/m²K⁴)
A is the skin area (given as 2.0 m²)
T is the skin temperature (34°C + 273.15 K = 307.15 K)
T₀ is the room temperature (25°C + 273.15 K = 298.15 K)

Plugging in the values into the formula:

Q/t = (0.80)(5.67 x 10^-8 W/m²K⁴)(2.0 m²)((307.15 K)² - (298.15 K)²)

Q/t = 0.80 x 5.67 x 10^-8 x 2.0 x (307.15² - 298.15²)

Q/t = 0.80 x 5.67 x 10^-8 x 2.0 x (94422.9225 - 88919.4225)

Q/t = 0.80 x 5.67 x 10^-8 x 2.0 x 5503.5

Q/t ≈ 0.793 W

So, the net loss of radiant power from the body is approximately 0.793 Watts.

To calculate the net loss of radiant power from the body, we can use the Stefan-Boltzmann law and the formula you provided for radiant power.

a) The formula you used is correct. However, I noticed a mistake in your calculation. The Stefan-Boltzmann constant is approximately 5.67 x 10^-8 W/(m^2K^4). Let's reconfigure the calculation with the correct value.

Q/t = emissivity x Stefan-Boltzmann constant x (T^4 - T0^4) x A

Where:
Q/t is the rate of heat transfer (in watts),
emissivity is the emissivity of the skin,
Stefan-Boltzmann constant is approximately 5.67 x 10^-8 W/(m^2K^4),
T is the temperature of the skin (in Kelvin),
T0 is the temperature of the surroundings (in Kelvin), and
A is the surface area of the skin.

Given values:
emissivity = 0.80,
Stefan-Boltzmann constant = 5.67 x 10^-8 W/(m^2K^4),
T = 34°C = 307.15 K (convert to Kelvin),
T0 = 25°C = 298.15 K (convert to Kelvin),
A = 2.0 m^2.

Now let's substitute the values and calculate:

Q/t = 0.8 x (5.67 x 10^-8) x (307.15^4 - 298.15^4) x 2.0

Please note that the temperatures need to be in Kelvin to maintain consistency with the units of the Stefan-Boltzmann constant.

After substituting the values and calculating, you should find the correct net loss of radiant power from the body in watts.

b) Once you have obtained the net loss rate of power from part (a) in watts, you can convert it to energy in joules per hour. As you correctly mentioned, 1 watt is equal to 3600 J/h (joules per hour).

To calculate the total energy loss in joules per hour, multiply the net loss rate of power (in watts) by 3600 J/h.

Finally, to convert the energy in joules to food Calories, divide the value by 4186 J, since 1 food Calorie (with a capital "C") is equal to 4186 J.

I hope this clarifies the procedure. Let me know if you have any further questions or need assistance with the calculations.