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 = 0.810). It has a temperature of 65.0 oC. The new owner of the house paints the radiator a lighter color (emissitivity = 0.414). 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?

Question

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 = 0.810). It has a temperature of 65.0 oC. The new owner of the house paints the radiator a lighter color (emissitivity = 0.414). 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?

Answer 1

To solve this problem, we'll use the Stefan-Boltzmann law, which relates the power radiated by an object to its temperature and emissivity.

The Stefan-Boltzmann law states that the power radiated per unit area (P) by an object is proportional to the fourth power of its temperature (T) and is given by the equation:

P = σ * ε * A * T^4

Where:
- P is the power radiated by the object,
- σ is the Stefan-Boltzmann constant (5.67 x 10^-8 W/m^2K^4),
- ε is the emissivity of the object,
- A is the surface area of the object, and
- T is the temperature of the object.

Now, let's consider the initial conditions where the radiator has a dark color with an emissivity of 0.810. We'll denote this as T1.

P1 = σ * ε1 * A * (T1)^4

Next, let's consider the new conditions where the radiator has been painted a lighter color with an emissivity of 0.414. We'll denote the temperature of the newly painted radiator as T2.

P2 = σ * ε2 * A * (T2)^4

The problem states that the radiator emits the same radiant power before and after being painted. Therefore, we can equate P1 and P2:

P1 = P2

σ * ε1 * A * (T1)^4 = σ * ε2 * A * (T2)^4

Since the surface area (A) and the Stefan-Boltzmann constant (σ) are common to both sides of the equation, we can eliminate them.

ε1 * (T1)^4 = ε2 * (T2)^4

Now, we can rearrange the equation to solve for T2:

(T2)^4 = (ε1 / ε2) * (T1)^4

Taking the fourth root of both sides, we have:

T2 = (ε1 / ε2)^(1/4) * T1

Substituting the given values, we get:

T2 = (0.810 / 0.414)^(1/4) * 65.0 oC

T2 ≈ 48.7 oC

Therefore, the temperature of the newly painted radiator is approximately 48.7 degrees Celsius.