A dish of hot food has an emissivity of 0.42 and emits 15 W of thermal radiation. If you wrap it in aluminum foil, which has an emissivity of 0.06, how much power will it radiate?
To determine the power that the dish of hot food will radiate after being wrapped in aluminum foil, you need to use the Stefan-Boltzmann Law. This law relates the power radiated by an object to its emissivity and temperature.
The Stefan-Boltzmann Law states that the power radiated by an object is proportional to the emissivity (ε) of the object, multiplied by the Stefan-Boltzmann constant (σ), multiplied by the temperature (T) raised to the fourth power.
Mathematically, the equation is expressed as:
P = ε * σ * T^4
In this case, you have two scenarios:
1. The dish of hot food without aluminum foil, with emissivity (ε₁ = 0.42) and power radiated (P₁ = 15 W).
2. The dish of hot food wrapped in aluminum foil, with aluminum foil emissivity (ε₂ = 0.06) and unknown power radiated (P₂).
Since both scenarios involve the same dish of hot food, the temperature (T) remains the same in both cases. Therefore, we can write the ratio of power radiated by the two scenarios as:
P₂ / P₁ = (ε₂ * σ * T^4) / (ε₁ * σ * T^4)
Canceling out the common factors, we get:
P₂ / P₁ = ε₂ / ε₁
To find P₂, rearrange the equation:
P₂ = P₁ * (ε₂ / ε₁)
Substituting the given values:
P₂ = 15 W * (0.06 / 0.42)
P₂ = 2.14 W
Therefore, when the dish of hot food is wrapped in aluminum foil, it will radiate approximately 2.14 Watts of thermal radiation.