A person sitting quietly in a room on a winter day may lose heat primarily by radiation from the outer surface of her clothes to the walls of the room. If the walls are at 11.5° C and the rate of radiation is 75.2 kcal/h (a typical basal metabolic rate), what is the temperature of the outer surface of the clothes, which have an emissivity of 0.800? (Assume the surface area of the person is 1 m2.)
To solve this problem, we can use the Stefan-Boltzmann Law, which relates the rate of radiation emitted by an object to its temperature and emissivity.
The Stefan-Boltzmann Law states that the rate of radiation (P) emitted by an object is given by the formula P = εσAT^4, where ε is the emissivity of the object, σ is the Stefan-Boltzmann constant (approximately 5.67 x 10^-8 W/m^2*K^4), A is the surface area of the object, and T is the temperature in Kelvin.
Here's how we can calculate the temperature of the outer surface of the clothes:
Step 1: Convert the rate of radiation from kcal/h to Watts.
75.2 kcal/h = 75.2 * 4.184 kJ/h (since 1 kcal = 4.184 kJ)
= 75.2 * 4.184 * 10^3 J/h (since 1 kJ = 10^3 J)
= 75.2 * 4.184 * 10^3 / 3.6 J/s (since 1 hour = 3.6 seconds)
≈ 87,053 J/s (Watts)
Step 2: Substitute the known values into the Stefan-Boltzmann Law.
87,053 = 0.800 * 5.67 x 10^-8 * 1 * T^4
Step 3: Solve for T.
T^4 = (87,053 / (0.800 * 5.67 x 10^-8))
T^4 ≈ 2,420,700,892
Taking the fourth root of both sides:
T ≈ (2,420,700,892)^(1/4)
T ≈ 101.7 K
Step 4: Convert the temperature back to Celsius.
T in Celsius = T in Kelvin - 273.15
T in Celsius ≈ 101.7 - 273.15
T in Celsius ≈ -171.45 °C
Therefore, the temperature of the outer surface of the clothes is approximately -171.45 °C.