A person's body is producing energy internally due to metabolic processes. If the body loses more energy than metabolic processes are generating, its temperature will drop. If the drop is severe, it can be life-threatening. Suppose a person is unclothed and energy is being lost via radiation from a body surface area of 1.56 m2, which has a temperature of 34 °C and an emissivity of 0.673. Suppose that metabolic processes are producing energy at a rate of 137 J/s. What is the temperature in kelvins of the coldest room in which this person could stand and not experience a drop in body temperature?
Net radiative heat loss
= sigma*A*[(307)^4 - T^4]
= 137 watts
"sigma" is the Stefan Boltzmann constant. Look it up if you have never heared of it. A is the body area.
An emissivity of 1 has been assumed. Convective heat loss has been ignored. These may not be accurate assumptions.
Solve for T, the room temperature in Kelvin.
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To solve this problem, we can use the concept of heat transfer and energy balance.
The energy lost by the body via radiation can be calculated using the Stefan-Boltzmann Law:
Energy lost (Q) = ε * σ * A * (T^4 - T_room^4)
Where:
ε = emissivity of the body's surface
σ = Stefan-Boltzmann constant (5.67 x 10^-8 W/m^2K^4)
A = body surface area
T = body temperature in Kelvin
T_room = room temperature in Kelvin
We want to find the coldest room temperature (T_room) that will not result in a drop in body temperature. In this case, the energy lost via radiation should be equal to the energy generated by metabolic processes:
Energy lost (Q) = Energy generated (W)
Replacing the variables with given values:
ε * σ * A * (T^4 - T_room^4) = 137 J/s
Given:
A = 1.56 m^2
ε = 0.673
T = 34 °C + 273.15 = 307.15 K
Now we can solve for T_room:
0.673 * (5.67 x 10^-8 W/m^2K^4) * (1.56 m^2) * (307.15 K^4 - T_room^4) = 137 J/s
Simplifying the equation:
T_room^4 = (0.673 * (5.67 x 10^-8) * (1.56) * (307.15^4) - 137) / (0.673 * (5.67 x 10^-8) * (1.56))
Calculating the right side of the equation:
T_room^4 = 235635.66
Taking the fourth root of both sides:
T_room = 22.86 K
Therefore, the coldest room temperature in which this person could stand without experiencing a drop in body temperature is approximately 22.86 Kelvin.
To determine the temperature in kelvins of the coldest room in which this person could stand without experiencing a drop in body temperature, we need to consider the heat exchanged between the person and the room.
The heat lost by the person's body through radiation can be calculated using the Stefan-Boltzmann law:
Heat lost (Q) = ε * σ * A * ΔT^4
Where:
ε = emissivity of the body surface (0.673)
σ = Stefan-Boltzmann constant (5.67 x 10^-8 W/m^2K^4)
A = surface area of the body (1.56 m^2)
ΔT^4 = temperature difference between the body and the environment in kelvins, raised to the power of 4 (T_body - T_room)^4
The heat generated by the metabolic processes is given as 137 J/s. The person will not experience a drop in body temperature if the heat generated equals the heat lost to the environment, i.e., Q = metabolic heat production.
Substituting the given values:
137 J/s = 0.673 * (5.67 x 10^-8 W/m^2K^4) * 1.56 m^2 * (T_body - T_room)^4
Simplifying the equation:
(T_body - T_room)^4 = (137 J/s) / (0.673 * (5.67 x 10^-8 W/m^2K^4) * 1.56 m^2)
Taking the fourth root of both sides:
T_body - T_room = ((137 J/s) / (0.673 * (5.67 x 10^-8 W/m^2K^4) * 1.56 m^2))^(1/4)
Now, let's calculate the value inside the parentheses on the right side of the equation:
((137 J/s) / (0.673 * (5.67 x 10^-8 W/m^2K^4) * 1.56 m^2))^(1/4) ≈ 30.77 K
Therefore:
T_body - T_room ≈ 30.77 K
Rearranging the equation to solve for T_room:
T_room = T_body - 30.77 K
Substituting the given body temperature (34 °C) into the equation:
T_room = (34 + 273.15) K - 30.77 K
T_room ≈ 276.38 K
Thus, the coldest room temperature in kelvins in which this person could stand without experiencing a drop in body temperature is approximately 276.38 K.