A sphere of radius 30cm, temperature 23.0°C and emissivity 0.62 is located in an environment of temperature 58.0°C. At what rate does the sphere emit and absorb thermal radiation and what's the sphere net rate of energy exchange?
To calculate the rate at which the sphere emits and absorbs thermal radiation, we can use Stefan-Boltzmann's Law and the formula for net rate of energy exchange.
1. Calculate the rate of thermal radiation emitted by the sphere using Stefan-Boltzmann's Law:
The formula for the rate of thermal radiation emitted is given by
P_emitted = ε * σ * A * (T_s^4 - T_env^4)
Where:
P_emitted = rate of thermal radiation emitted
ε = emissivity of the sphere
σ = Stefan-Boltzmann constant (5.67 x 10^-8 W/m^2K^4)
A = surface area of the sphere
T_s = temperature of the sphere
T_env = temperature of the environment
Substitute the given values:
ε = 0.62
σ = 5.67 x 10^-8 W/m^2K^4
A = 4πr^2 = 4π(0.3)^2
T_s = 23.0 + 273.15 (convert °C to Kelvin)
T_env = 58.0 + 273.15 (convert °C to Kelvin)
Calculate the values and solve for P_emitted.
2. Calculate the rate of thermal radiation absorbed by the sphere:
The rate of thermal radiation absorbed is equal to the rate of thermal radiation emitted since they are in thermal equilibrium.
P_absorbed = P_emitted
3. Calculate the net rate of energy exchange:
The net rate of energy exchange is given by
Net_rate = P_emitted - P_absorbed
Substitute the previously calculated values and solve for Net_rate.
Let's calculate the values step-by-step.
To calculate the rate at which the sphere emits and absorbs thermal radiation, as well as the net rate of energy exchange, you can use the Stefan-Boltzmann Law and the equation for net energy exchange.
According to the Stefan-Boltzmann Law, the rate at which an object emits thermal radiation (P_emission) is given by:
P_emission = ε * σ * A * (T_sphere^4 - T_env^4)
where:
- ε is the emissivity of the sphere (0.62 in this case)
- σ is the Stefan-Boltzmann constant (5.67 x 10^-8 W/m^2K^4)
- A is the surface area of the sphere (4π * r^2, where r is the radius of the sphere)
- T_sphere is the temperature of the sphere (23.0°C, which needs to be converted to Kelvin by adding 273.15)
- T_env is the temperature of the environment (58.0°C, also converted to Kelvin)
Now, let's calculate the values step by step:
1. Convert the temperatures to Kelvin:
T_sphere = 23.0°C + 273.15 = 296.15 K
T_env = 58.0°C + 273.15 = 331.15 K
2. Calculate the surface area of the sphere:
A = 4π * r^2 = 4π * (30 cm)^2 = 4π * (0.3 m)^2 = 4π * 0.09 m^2 = 0.113 m^2
3. Calculate the rate of emission:
P_emission = 0.62 * (5.67 x 10^-8 W/m^2K^4) * 0.113 m^2 * (296.15 K^4 - 331.15 K^4)
P_emission = 2.012 x 10^-5 W
Now, to calculate the rate at which the sphere absorbs thermal radiation, we assume that the sphere is a perfect absorber, so the rate of absorption (P_absorption) will be equal to the rate of emission (P_emission):
P_absorption = P_emission = 2.012 x 10^-5 W
Finally, to determine the net rate of energy exchange, we subtract the rate of emission from the rate of absorption:
Net rate of energy exchange = P_absorption - P_emission
= 2.012 x 10^-5 W - 2.012 x 10^-5 W
= 0
Therefore, the net rate of energy exchange is 0.