A naked person (e=0.75) with a surface area 2.00 square meters and a skin temperature of 34 degrees Celsius is in sauna at 90 degrees Celsius. If he stays in the sauna for 30 minutes, (a) How much heat does the perspn radiates? (b) How much heat does he absorb from the walls of the sauna assuming that the wall is a black body?
9, I think
I am not good at physics, sooooo...
I got 29% in my first test so I tried hard and ended up with 100 !!!
so the answer is 9
hope that answered your question!¬
To calculate the heat radiated by the person and the heat absorbed from the sauna walls, we can use the Stefan-Boltzmann Law. The formula is as follows:
Q = εσA(T1^4 - T2^4)
Where:
Q is the heat radiated or absorbed
ε is the emissivity of the surface (0.75 for a naked person)
σ is the Stefan-Boltzmann constant (approximately 5.67 x 10^-8 W/(m^2·K^4))
A is the surface area of the person (2.00 square meters)
T1 is the temperature of the person's skin (34 degrees Celsius + 273.15 to convert to Kelvin)
T2 is the temperature of the sauna (90 degrees Celsius + 273.15 to convert to Kelvin)
(a) To calculate the heat radiated by the person, we use the formula with the person's surface area, emissivity, and temperature:
Q_radiated = εσA(T1^4 - T2^4)
Substituting the values:
Q_radiated = 0.75 * 5.67x10^-8 * 2.00 * (307.15^4 - 363.15^4)
Q_radiated ≈ 2021.25 Watts
Therefore, the person radiates approximately 2021.25 Watts of heat.
(b) To calculate the heat absorbed from the sauna walls, we use the same formula but switch the temperatures:
Q_absorbed = εσA(T2^4 - T1^4)
Substituting the values:
Q_absorbed = 0.75 * 5.67x10^-8 * 2.00 * (363.15^4 - 307.15^4)
Q_absorbed ≈ 10077.36 Watts
Therefore, the person absorbs approximately 10077.36 Watts of heat from the sauna walls.