A copper ball 2 cm in radius is heated in a furnace to 400 degree celsius. if its emissivity is 0.3, at what rate does it radiates?
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To calculate the rate at which the copper ball radiates heat, we can use the Stefan-Boltzmann law. The Stefan-Boltzmann law states that the power radiated by an object is directly proportional to the fourth power of its absolute temperature and its emissivity.
The formula for calculating the rate at which an object radiates heat is:
P = ε * σ * A * T^4
Where:
P is the power radiated (in watts)
ε is the emissivity of the material (given as 0.3)
σ is the Stefan-Boltzmann constant (approximately 5.67 × 10^-8 W/m^2K^4)
A is the surface area of the object (4πr^2, where r is the radius)
T is the absolute temperature (in Kelvin)
To convert the Celsius temperature to Kelvin, we need to add 273.15 to the given Celsius temperature (400°C).
Let's calculate the rate at which the copper ball radiates heat:
Step 1: Convert the Celsius temperature to Kelvin.
T = 400 + 273.15
T = 673.15 K
Step 2: Calculate the surface area of the copper ball.
A = 4πr^2
Given radius (r) = 2 cm = 0.02 m (convert to meters)
A = 4 * 3.14159 * (0.02 m)^2
Step 3: Substitute the values into the formula and calculate the power radiated (P).
P = 0.3 * (5.67 × 10^-8 W/m^2K^4) * [4 * 3.14159 * (0.02 m)^2] * (673.15 K)^4
Using the appropriate unit conversions and performing the calculations gives us the power radiated (P) by the copper ball.