A sphere of radius 0.583 m, temperature 39.4°C, and emissivity 0.983 is located in an environment of temperature 85.1°C. At what rate does the sphere (a) emit and (b) absorb thermal radiation? (c) What is the sphere's net rate of energy exchange?
The area of the sphere is:
A = 4 pi R^2
where R is the radius
The emitted power is given by:
P_em = emissivity*A*sigma T1^4
where T1 is the absolute temperature of the sphere
The absorbed power is given by:
P_ab = emissivity*A*sigma T2^4
where T2 is the absolute temperature of the environment.
sigma is given by:
sigma = 2 pi^5 k^4/(15 h^3 c^2)
Here k is Boltzmann's constant, h is Planck's constant and c is the speed of light. sigma is approximately given by:
sigma = 5.67*10^(-8) Watt/(K^4 m^2)
To calculate the rate at which the sphere emits and absorbs thermal radiation, we can use the Stefan-Boltzmann Law, which states that the rate of thermal radiation emitted by an object is proportional to the fourth power of its temperature (in Kelvin) and surface area, and is also affected by its emissivity.
Let's break down each part of the problem:
(a) Rate of Thermal Radiation Emitted:
To calculate the rate at which the sphere emits thermal radiation, we can use the formula:
E_emit = ε * σ * A * (T_sphere^4 - T_env^4)
Where:
E_emit = Rate of thermal radiation emitted
ε = Emissivity of the sphere
σ = Stefan-Boltzmann constant (approximated to 5.67 * 10^-8 W/(m^2*K^4))
A = Surface area of the sphere
T_sphere = Temperature of the sphere (in Kelvin)
T_env = Temperature of the environment (in Kelvin)
First, let's convert the temperatures to Kelvin:
T_sphere = 39.4°C + 273.15 = 312.55 K
T_env = 85.1°C + 273.15 = 358.25 K
Now, calculate the surface area of the sphere:
The surface area of a sphere is given by the formula: A = 4πr^2
where r is the radius of the sphere.
A = 4π * (0.583 m)^2
Now, substitute the values into the formula:
E_emit = 0.983 * 5.67 * 10^-8 * 4π * (0.583 m)^2 * (312.55 K^4 - 358.25 K^4)
Calculate this value to find the rate at which the sphere emits thermal radiation.
(b) Rate of Thermal Radiation Absorbed:
The rate at which the sphere absorbs thermal radiation is equal to the rate at which it emits thermal radiation since it is in thermal equilibrium with its environment. Therefore, the rate of thermal radiation absorbed by the sphere is the same as the rate of thermal radiation emitted.
E_absorb = E_emit (from part a)
(c) Net Rate of Energy Exchange:
The net rate of energy exchange is the difference between the rates of thermal radiation absorbed and emitted by the sphere.
Net rate = E_absorb - E_emit (from part b and a)
Now, calculate the net rate of energy exchange to get the final answer.
By following these steps and substituting the values into the formulas, you'll be able to determine the rate at which the sphere emits, absorbs, and the net rate of energy exchange.