# The surface area of an adult human is about 1.8m2 . Suppose a person with a skin temperature of 34 ∘C is standing in a room where the air is 25∘C but the walls are 17 ∘C.

A) There is a "dead-air" layer next to your skin that acts as insulation. If the dead-air layer is 5.0mm thick, what is the person's rate of heat loss by conduction?

B) What is the person's net radiation loss to the walls? The emissivity of skin is 0.97.

## A) To calculate the rate of heat loss by conduction, we can use the formula:

Q = k * A * ΔT / d

Where:
Q is the rate of heat loss by conduction,
k is the thermal conductivity of air,
A is the surface area of the person,
ΔT is the temperature difference between the person's skin and the air,
d is the thickness of the dead-air layer.

First, let's find the temperature difference (ΔT) between the person's skin and the air:
ΔT = skin temperature - air temperature
= 34°C - 25°C
= 9°C

The thermal conductivity of air (k) is approximately 0.024 W/(m·K).

The surface area of an adult human is given as 1.8 m².

The thickness of the dead-air layer (d) is 5.0 mm, which is equal to 0.005 m.

Now, we can substitute these values into the formula and calculate the rate of heat loss by conduction:

Q = (0.024 W/(m·K)) * (1.8 m²) * (9°C) / (0.005 m)
≈ 77.76 W

Therefore, the person's rate of heat loss by conduction is approximately 77.76 W.

B) The net radiation loss to the walls can be calculated using the formula:

Q = ε * σ * A * (T₁⁴ - T₂⁴)

Where:
Q is the rate of radiation loss,
ε is the emissivity of the skin (given as 0.97),
σ is the Stefan-Boltzmann constant (approximately 5.67 × 10^(-8) W/(m²·K⁴)),
A is the surface area of the person,
T₁ is the person's skin temperature in Kelvin,
T₂ is the wall temperature in Kelvin.

First, we need to convert the temperatures into Kelvin:
T₁ = skin temperature + 273.15
= 34°C + 273.15
= 307.15 K

T₂ = wall temperature + 273.15
= 17°C + 273.15
= 290.15 K

Now, we can substitute these values into the formula and calculate the net radiation loss to the walls:

Q = (0.97) * (5.67 × 10^(-8) W/(m²·K⁴)) * (1.8 m²) * (307.15 K⁴ - 290.15 K⁴)
≈ 30.54 W

Therefore, the person's net radiation loss to the walls is approximately 30.54 W.

## A) To find the person's rate of heat loss by conduction, we can use the formula for heat conduction:

Q = k * A * ΔT / d

Where:
- Q is the heat transfer rate (in watts),
- k is the thermal conductivity of the material (in watts per meter per degree Celsius),
- A is the surface area (in square meters),
- ΔT is the temperature difference (in degrees Celsius), and
- d is the thickness of the material (in meters).

In this case, the thermal conductivity of dead air is approximately 0.024 W/(m⋅°C) and the thickness of the dead-air layer is 5.0 mm (or 0.005 meters).

The temperature difference is the difference between the skin temperature and the air temperature, so ΔT = 34°C - 25°C = 9°C.

Plugging in these values, the equation becomes:

Q = 0.024 * 1.8 * 9 / 0.005

Simplifying the formula:

Q = 77.76 watts

Therefore, the person's rate of heat loss by conduction is approximately 77.76 watts.

B) To calculate the person's net radiation loss to the walls, we can use the Stefan-Boltzmann law:

Q = ε * σ * A * (T1^4 - T2^4)

Where:
- Q is the heat transfer rate (in watts),
- ε is the emissivity (a dimensionless value ranging from 0 to 1, where 0 represents a perfect reflector and 1 represents a perfect absorber),
- σ is the Stefan-Boltzmann constant, approximately equal to 5.67 × 10^-8 W/(m^2⋅K^4),
- A is the surface area (in square meters),
- T1 is the temperature of the person (in Kelvin), and
- T2 is the temperature of the walls (in Kelvin).

Given that the emissivity of skin is 0.97, the surface area of an adult human is 1.8 m^2, the person's skin temperature is 34°C (or 307.15 K), and the wall temperature is 17°C (or 290.15 K), the equation becomes:

Q = 0.97 * 5.67 × 10^-8 * 1.8 * (307.15^4 - 290.15^4)

Simplifying the formula:

Q ≈ 193.24 watts

Therefore, the person's net radiation loss to the walls is approximately 193.24 watts.