A sphere of radius 0.715 m, temperature 21.1°C, and emissivity 0.672 is located in an environment of temperature 93.8°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?
To determine the rate at which the sphere emits thermal radiation, we can use the Stefan-Boltzmann law, which states that the power radiated by a black body (or an object with a known emissivity) is proportional to the fourth power of its temperature.
(a) To calculate the rate at which the sphere emits thermal radiation, we can use the following equation:
P_emit = ε * σ * A * (T_sphere^4 - T_env^4),
where:
P_emit is the rate at which the sphere emits thermal radiation,
ε is the emissivity of the sphere (given as 0.672),
σ is the Stefan-Boltzmann constant (approximately 5.67 * 10^-8 W/(m^2*K^4)),
A is the surface area of the sphere (given by 4πr^2), and
T_sphere and T_env are the temperatures of the sphere and the environment, respectively.
Substituting the given values into the equation, we have:
P_emit = 0.672 * 5.67 * 10^-8 * 4π * (0.715^2) * (294.1^4 - 366.9^4),
Now we can calculate the rate at which the sphere emits thermal radiation.
(b) To calculate the rate at which the sphere absorbs thermal radiation, we can use the following equation:
P_absorb = ε * σ * A * (T_env^4 - T_sphere^4).
Substituting the given values into the equation, we have:
P_absorb = 0.672 * 5.67 * 10^-8 * 4π * (0.715^2) * (366.9^4 - 294.1^4).
Now we can calculate the rate at which the sphere absorbs thermal radiation.
(c) The net rate of energy exchange is determined by subtracting the rate of absorption from the rate of emission:
P_net = P_emit - P_absorb.
By substituting the previously calculated values, we can find the net rate of energy exchange.