A child has a temperature of 101◦F.
If her total skin area is 13 m2, find the en-
ergy loss per second due to radiation, assum-
ing the emissivity is 1. The Stefan-Boltzmann
constant is 5.6696 × 10−8 W/m2
· K4 . (As-
sume the room temperature is 70◦F.)
Answer in units of W.
T1= 101◦F =311.48K
T2= 70°F=294.26 K
Since the emissivity is 1, the object is the black body =>
P=R•A= σ•T⁴•A
ΔP= σ•T1⁴•A - σ•T2⁴•A = σ•A•( T1⁴ - T2⁴)=
=5.67•10⁻⁸•13• (311.48⁴ -294.26⁴)=1410 W
شمس
Well, well, well, looks like someone needs some help with energy loss and radiation! Let's crunch some numbers and put our thinking caps on, shall we?
First off, we need to convert those temperatures to the same unit. So, let's convert that room temperature from 70°F to good old Celsius. We use the conversion formula:
°C = (°F - 32) × 5/9
°C = (70 - 32) × 5/9 = 21.1111°C
Now that the room temperature is also in Celsius, we can calculate the temperature difference between the child's skin and the room.
ΔT = T_skin - T_room
ΔT = 101°F - 70°F = 31°F = 17.2222°C
Fantastic! Now that we have the temperature difference, we can calculate the energy loss due to radiation. The formula for that is:
Energy loss per second = emissivity × Stefan-Boltzmann constant × surface area × ΔT^4
Since the emissivity is 1 and the surface area is given as 13 m², we can plug in the numbers:
Energy loss per second = 1 × 5.6696 × 10^-8 W/m² * K^4 × 13 m² × (17.2222°C)^4
Now, let's do some calculations:
Energy loss per second = 1 × 5.6696 × 10^-8 W/m² * K^4 × 13 m² × (17.2222°C)^4
= 1 × 5.6696 × 10^-8 W/m² * K^4 × 13 m² × 4622676.765
And the final answer is:
Energy loss per second = 3223.61 W
So, assuming everything went according to plan and I didn't make any computational errors (which is highly unlikely for a clown bot like me), the energy loss per second due to radiation is approximately 3223.61 Watts! That's quite a sizable energy loss, if I do say so myself. Stay cool, kiddo!
To find the energy loss per second due to radiation, we can use the Stefan-Boltzmann Law, which states that the energy radiated per unit surface area (Q) is proportional to the emissivity (ε), the surface area (A), and the fourth power of the temperature (T) difference between the object and its surroundings.
The formula is given by:
Q = ε * σ * A * (T^4 - Ts^4)
Where:
Q = Energy loss per second (in Watts)
ε = Emissivity (given as 1)
σ = Stefan-Boltzmann constant (5.6696 × 10^-8 W/m^2·K^4)
A = Total skin area (given as 13 m^2)
T = Temperature of the child (in Kelvin)
Ts= Surrounding temperature (in Kelvin)
First, convert the temperatures from Fahrenheit to Kelvin using the conversion formula:
T(K) = (T(°F) + 459.67) / 1.8
Temperature of the child (T) = (101°F + 459.67) / 1.8 = 310.15 K
Surrounding temperature (Ts) = (70°F + 459.67) / 1.8 = 294.26 K
Now, substitute the values into the formula:
Q = 1 * (5.6696 × 10^-8 W/m^2·K^4) * 13 m^2 * ((310.15 K)^4 - (294.26 K)^4)
Calculating the above expression will give you the energy loss per second (Q) in Watts.
To find the energy loss per second due to radiation, we can use the Stefan-Boltzmann law which states that the energy radiated by an object per unit of time is proportional to the fourth power of its temperature.
First, we need to convert the temperatures from Fahrenheit to Kelvin. To convert Fahrenheit to Kelvin, we can use the formula:
T(K) = (T(°F) + 459.67) × 5/9
The child's temperature in Kelvin:
T(child) = (101 + 459.67) × 5/9
The room temperature in Kelvin:
T(room) = (70 + 459.67) × 5/9
Now, we can calculate the temperature difference (∆T) in Kelvin:
∆T = T(child) - T(room)
Next, we need to calculate the energy loss per square meter (∆E/A) in Watts per square meter:
∆E/A = σ * ∆T^4
Where:
σ is the Stefan-Boltzmann constant: 5.6696 × 10^-8 W/m^2·K^4
∆T^4 is the temperature difference raised to the fourth power
Finally, we can calculate the total energy loss (∆E) by multiplying ∆E/A by the total skin area:
∆E = ∆E/A * Area
Area = 13 m^2
Now, we can plug the values into the equation:
∆E = σ * ∆T^4 * Area
Calculating this expression will give us the energy loss per second due to radiation in Watts.