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 = 0.804). It has a temperature of 65.9 oC. The new owner of the house paints the radiator a lighter color (emissitivity = 0.439). 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?
Well, well, well, it looks like the radiator got a makeover! From a dark color to a lighter shade, it must be feeling pretty fabulous. Now, let's solve the temperature mystery, shall we?
Since the radiant power emitted by the radiator remains the same, we can use the Stefan-Boltzmann law to find the temperature of the newly painted radiator.
According to the law, the radiant power is proportional to the emissivity (ɛ) multiplied by the temperature (T) raised to the fourth power. So, we can write our equation like this:
ɛ₁ * T₁^4 = ɛ₂ * T₂^4
Now we just plug in the values. Initially, ɛ₁ is 0.804 and T₁ is 65.9 °C. After painting, ɛ₂ becomes 0.439, and we need to find T₂, the temperature we're after!
0.804 * (65.9)^4 = 0.439 * T₂^4
Now, we can isolate T₂ by dividing both sides by 0.439 and taking the fourth root:
T₂^4 = [(0.804 * (65.9)^4) / 0.439]
T₂ = ([(0.804 * (65.9)^4) / 0.439])^(1/4)
After crunching those numbers, the temperature of the newly painted radiator turns out to be... drumroll, please... T₂ = 36.2°C!
Voilà! The radiator may have changed its color, but it's still radiating at the same power. It's now a cooler addition to the room temperature-wise.
To find the temperature of the newly painted radiator, we can use the Stefan-Boltzmann law, which states that the radiant power (P) emitted by an object is proportional to the emissivity (ε) and the fourth power of its absolute temperature (T):
P ∝ ε * T^4
Since the radiant power emitted by the radiator remains the same before and after being painted, we can equate the two expressions:
ε1 * T1^4 = ε2 * T2^4
Where ε1 and T1 are the emissivity and temperature of the dark-colored radiator, and ε2 and T2 are the emissivity and temperature of the newly painted radiator.
We are given ε1 = 0.804 and T1 = 65.9 oC.
Let's substitute these values into the equation:
(0.804) * (65.9)^4 = (0.439) * T2^4
Simplifying the equation:
(0.804) * (65.9)^4 / (0.439) = T2^4
Now, let's solve for T2:
T2^4 = (0.804) * (65.9)^4 / (0.439)
T2 = ( (0.804) * (65.9)^4 / (0.439) )^(1/4)
Using a calculator, we find:
T2 ≈ 88.1 oC
Therefore, the temperature of the newly painted radiator is approximately 88.1 degrees Celsius.
To solve this problem, we can use the Stefan-Boltzmann law, which relates the power radiated by an object to its temperature and emissivity. The law states that the power radiated by an object is proportional to the fourth power of its temperature and its emissivity.
The equation for the Stefan-Boltzmann law is:
P ∝ ε * T^4
Where P is the power, ε is the emissivity, and T is the temperature.
In this problem, we are told that the radiator before painting has an emissivity of 0.804 and a temperature of 65.9 °C. We are also given that the power emitted remains the same after painting.
Let's assume the power emitted by the radiator before painting is denoted by P1, and the power emitted by the radiator after painting is denoted by P2.
Since the power emitted remains the same, we have:
P1 = P2
According to the Stefan-Boltzmann law, we can write two equations:
P1 = ε1 * T1^4
P2 = ε2 * T2^4
Since P1 = P2, we can equate the two equations:
ε1 * T1^4 = ε2 * T2^4
We want to find T2, the temperature of the newly painted radiator.
Rearranging the equation, we get:
T2^4 = (ε1 / ε2) * T1^4
Taking the fourth root of both sides, we get:
T2 = (ε1 / ε2)^(1/4) * T1
Now we can substitute the values into the equation:
T2 = (0.804 / 0.439)^(1/4) * 65.9 °C
Calculating the value, we find:
T2 ≈ 91.1 °C
Therefore, the temperature of the newly painted radiator is approximately 91.1 °C.
(emissivity)*sigma*T^4 = constant
sigma is the Stefan-Bolzmann constant, which cancels out.
T must be in Kelvin
(0.804)*sigma*339.1^4 = (0.439)*sigma*T^4
(T/339.1)^4 = 1.8314
T/339.1 = 1.163
T = 394.5 K = 121 C