Final answer:
The net rate of radiation Q_net from a pot at 41.9°C to a -24.1°C freezer, with a surface area of 0.133m² and emissivity of 0.683, is approximately 2.2662 x 10^7 W.
Step-by-step explanation:
The net rate of radiation can be calculated using the Stefan-Boltzmann Law, which is given by the formula:
Net rate of radiation = ε σ A (T_1^4 - T_2^4)
where:
- ε is the emissivity of the surface,
- σ is the Stefan-Boltzmann constant 5.67 x 10^{-8} W m}^{-2} K^{-4},
- A is the surface area,
- T_1 is the temperature of the warmer object (in Kelvin),
- T_2 is the temperature of the colder object (in Kelvin).
First, we need to convert the temperatures to Kelvin:
T_1 = 273.15 + 41.9 K
T_2 = 273.15 - 24.1 K
Now, we can substitute these values into the formula:
Net rate of radiation = (0.683) x (5.67 x 10^{-8}) x (0.133) x (273.15 + 41.9)^4 - (273.15 - 24.1)^4
Calculating this expression will give you the net rate of radiation.
Let's calculate the net rate of radiation using the provided values:
T_1 = 273.15 + 41.9 K = 315.05 K
T_2 = 273.15 - 24.1 K = 249.05 K
Now, substitute these values into the formula:
Net rate of radiation = (0.683) x (5.67 x 10^{-8}) x (0.133) x (315.05)^4 - (249.05)^4
Calculating this expression will give you the net rate of radiation. Let's compute it:
Net rate of radiation ≈ (0.683) x (5.67 x 10^{-8}) x (0.133) x (315.05)^4 - (249.05)^4
Net rate of radiation ≈ 0.683 x 5.67 x 10^{-8} x 0.133 x (7.169 x10^{14} - 2.552 x 10^{14})
Net rate of radiation ≈ 0.683 x 5.67 x 10^{-8} x 0.133 x 4.617 x 10^{14}
Net rate of radiation} ≈ 2.2662 x 10^7 W
So, the net rate of radiation by the pot is approximately 2.2662 x 10^7 W.