An air-cooled motorcycle engine loses a significant amount of heat through thermal radiation according to the Stefan-Boltzmann equation. Assume that the ambient temperature is T0 = 24.5°C (297.65 K). Suppose the engine generates 11.5 hp (8.58 kW) of power and, due to several deep surface fins, has a surface area of A = 0.540 m2. A shiny engine has an emissivity e = 0.0550, whereas an engine that is painted black has e = 0.950.

a) Determine the equilibrium temperature (in K) for the shiny engine. (Assume that radiation is the only mode by which heat is dissipated from the engine.)
b) And determine the equilibrium temperature (in K) for the black engine.

To determine the equilibrium temperature of the shiny engine, we can use the Stefan-Boltzmann equation:

P = εσAT^4

where P is the power generated by the engine, ε is the emissivity of the engine, σ is the Stefan-Boltzmann constant (5.67 x 10^-8 W/m^2K^4), A is the surface area of the engine, and T is the temperature of the engine in Kelvin.

a) For the shiny engine:

P = 8.58 kW = 8,580 W
ε = 0.0550
A = 0.540 m^2
σ = 5.67 x 10^-8 W/m^2K^4

Rearranging the equation to solve for T:

T^4 = P / (εσA)
T^4 = 8,580 / (0.0550 * 5.67 x 10^-8 * 0.540)
T^4 = 3.788 x 10^14
T ≈ (3.788 x 10^14)^(1/4)
T ≈ 626.2 K

Therefore, the equilibrium temperature of the shiny engine is approximately 626.2 Kelvin.

b) For the black engine:

P = 8.58 kW = 8,580 W
ε = 0.950
A = 0.540 m^2
σ = 5.67 x 10^-8 W/m^2K^4

Using the same equation:

T^4 = P / (εσA)
T^4 = 8,580 / (0.950 * 5.67 x 10^-8 * 0.540)
T^4 = 3.035 x 10^13
T ≈ (3.035 x 10^13)^(1/4)
T ≈ 515.1 K

Therefore, the equilibrium temperature of the black engine is approximately 515.1 Kelvin.