How Much Energy Does a Human Lose Through Radiation in a Cool Room?

[owura143]

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

 

[marcusl]

To get the net power loss from the body, you need to compute the power radiated by the warm skin, then subtract the power radiated by the room that is absorbed by the skin. Can you write the equation for that?

BTW, not sure why you used 25.15C degrees instead of 25C in your calc...

 

[m2003]

. Please double check your calculations and make sure to use the correct units. Additionally, it is important to consider the difference between radiant power and heat loss. Radiant power is the amount of energy emitted in the form of electromagnetic radiation, while heat loss is the transfer of thermal energy from the body to its surroundings. The net loss of radiant power can be calculated using the Stefan-Boltzmann law, but this does not necessarily equate to the total heat loss from the body. It is also important to consider other factors such as convection and conduction in determining the overall heat loss from a person's body.

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

Q/t = εσAT^4

where Q/t is the net loss of radiant power, ε is the emissivity, σ is the Stefan-Boltzmann constant (5.67 x 10^-8 W/m^2K^4), A is the surface area of the body (in this case, 2.0 m^2), and T is the temperature in Kelvin (34°C is equivalent to 307.15 K).

Plugging in these values, we get:

Q/t = (0.80)(5.67 x 10^-8 W/m^2K^4)(2.0 m^2)(307.15 K)^4 = 353.48 W

Therefore, the net loss of radiant power from the body is 353.48 watts.

b) To determine the number of food Calories of energy lost in one hour, we need to convert the radiant power to joules per hour:

353.48 watts x 3600 seconds = 1,272,528 joules/hour

To convert this to food Calories, we divide by 4186 J/Cal:

1,272,528 J/hour / 4186 J/Cal = 304.1 Calories/hour

Therefore, the person would need to consume approximately 304.1 Calories of food per hour to replace the energy lost through radiant power. However, as mentioned before, this does not take into account other sources of heat loss from the body, so the actual number of Calories needed may be higher.