In an old house, the heating system uses radiators, which are hollow metal devices through which hot water or steam circulates. In one room the radiator has a dark color (emissitivity = 0.818). It has a temperature of 64.8 oC. The new owner of the house paints the radiator a lighter color (emissitivity = 0.417). Assuming that it emits the same radiant power as it did before being painted, what is the temperature (in degrees Celsius) of the newly painted radiator?
To solve this problem, we can use the Stefan-Boltzmann law, which states that the power radiated by an object is proportional to its emissivity and temperature raised to the fourth power.
Let's denote the power radiated by the radiator before it was painted as P1 and after it was painted as P2.
According to the problem, the power radiated by the radiator before it was painted is the same as the power radiated after it was painted. Mathematically, we can write it as:
P1 = P2
Using the Stefan-Boltzmann law, we can express the power radiated as:
P = epsilon * sigma * T^4
Where:
- P is the power radiated
- epsilon is the emissivity of the radiator
- sigma is the Stefan-Boltzmann constant (approximately 5.67 x 10^-8 W/m^2K^4)
- T is the temperature of the radiator
Substituting the values for the emissivity (epsilon) and temperature (T) of the radiator before it was painted:
P1 = 0.818 * sigma * (64.8 + 273.15)^4
Substituting the values for the emissivity (epsilon) and temperature (T) of the newly painted radiator:
P2 = 0.417 * sigma * (T2 + 273.15)^4
Since P1 = P2, we can set the two equations equal to each other and solve for the temperature (T2) of the newly painted radiator:
0.818 * sigma * (64.8 + 273.15)^4 = 0.417 * sigma * (T2 + 273.15)^4
Simplifying the equation:
(64.8 + 273.15)^4 = (T2 + 273.15)^4
Taking the fourth root of both sides:
64.8 + 273.15 = T2 + 273.15
Simplifying further:
64.8 = T2
Therefore, the temperature (T2) of the newly painted radiator is 64.8 degrees Celsius.
To find the temperature of the newly painted radiator, we can use the Stefan-Boltzmann law, which states that the radiant power emitted by an object is proportional to its emissivity and temperature to the power of four.
The formula is: P = εσA(T^4)
Where:
P is the radiant power emitted.
ε is the emissivity.
σ is the Stefan-Boltzmann constant (5.67 x 10^(-8) W⋅m^(-2)⋅K^(-4)).
A is the surface area of the radiator.
T is the temperature of the radiator in Kelvin.
Since the radiant power emitted by the radiator remains the same, we can equate the two expressions:
P1 = P2
ε1σA(T1^4) = ε2σA(T2^4)
Since the surface area and constants are the same, we can cancel them out:
ε1(T1^4) = ε2(T2^4)
Now we can substitute the emissivities:
0.818 * (T1^4) = 0.417 * (T2^4)
Next, we rearrange the equation to solve for T2:
T2^4 = (0.818/0.417) * T1^4
T2 = (0.818/0.417)^(1/4) * T1
Finally, we plug in the values and calculate T2:
T2 = (0.818/0.417)^(1/4) * 64.8
T2 ≈ 85.9 oC
Therefore, the temperature of the newly painted radiator is approximately 85.9 oC.