An unclothed student is in 24


C room.
If the skin temperature of the student is
37

C, how much heat is lost from his body
in 6 min, assuming that the emissivity of the
skin is 0.83 and the surface area of the student
is 1.5 m2
. The Stephan-Boltzmann constant
is 5.6696 × 10
−8 W/m2
· K
4
.

To find the amount of heat lost from the student's body, we can use the Stefan-Boltzmann law. The Stefan-Boltzmann law states that the total power radiated by a body is directly proportional to the fourth power of its temperature and the surface area, and inversely proportional to the emissivity of the body.

The equation for heat loss by radiation is given by:

Q = εσAT^4

Where:
Q = Heat loss (in watts)
ε = Emissivity of the body (given as 0.83)
σ = Stefan-Boltzmann constant (given as 5.6696 × 10^-8 W/m^2·K^4)
A = Surface area of the student's body (given as 1.5 m^2)
T = Temperature of the student's body (given as 37°C)

To proceed with the calculation, we need to convert the temperature from Celsius to Kelvin:

T(K) = T(°C) + 273.15

T(K) = 37 + 273.15
T(K) = 310.15 K

Now, substitute the given values into the equation:

Q = (0.83)(5.6696 × 10^-8)(1.5)(310.15)^4

Using a calculator, calculate the value of Q.

Q = (0.83)(5.6696 × 10^-8)(1.5)(310.15^4)
Q ≈ 487.85 watts

Therefore, the student loses approximately 487.85 watts of heat from their body in 6 minutes in a 24°C room.