Question

Consider a circular grill whose diameter is \(0.3 \mathrm{~m}\). The bottom of the grill is covered with hot coal bricks at \(950 \mathrm{~K}\), while the wire mesh on top of the grill is covered with steaks initially at \(5^{\circ} \mathrm{C}\). The distance between the coal bricks and the steaks is \(0.20 \mathrm{~m}\). Treating both the steaks and the coal bricks as blackbodies, determine the initial rate of radiation heat transfer from the coal bricks to the steaks. Also, determine the initial rate of radiation heat transfer to the steaks if the side opening of the grill is covered by aluminum foil, which can be approximated as a reradiating surface.

Short answer

Answer: The initial rate of radiation heat transfer from the coal bricks to the steaks without the aluminum foil cover is approximately 5073.87 W, and with the aluminum foil cover, it is approximately 5327.56 W.

Step-by-step solution

Calculate the area of the coal bricks and steaks

First, we need to find the area of the circular grill, which represents the area of the coal bricks as well as the steaks. The area of a circle is given by \(A = \pi r^2\), where \(r\) is the radius of the circle. We're given the diameter \(d = 0.3 \mathrm{~m}\) so the radius will be \(r = \frac{d}{2} = 0.15 \mathrm{~m}\) Now, we can find the area: \(A = \pi (0.15)^2 \approx 0.0707 \mathrm{~m^2}\)

Convert temperature values to Kelvin

Before proceeding with the calculations, let's convert the temperature of the steaks from Celsius to Kelvin. We use the conversion formula \(T_{K} = T_{C} + 273.15\) \(T_{steaks} = 5^{\circ} \mathrm{C} + 273.15 = 278.15 \mathrm{K}\)

Calculate the initial rate of radiation heat transfer without aluminum foil

Now, we use the Stefan-Boltzmann Law to calculate the initial rate of radiation heat transfer from the coal bricks to the steaks considering both as blackbodies. The formula is \(q = A \sigma F_{A\to B}(T_A^4 - T_B^4)\), where \(A\) is the area, \(\sigma\) is the Stefan-Boltzmann constant, \(F_{A\to B}\) is the view factor from the coal bricks to the steaks, and \(T_A\) and \(T_B\) are the temperatures of the coal bricks and steaks, respectively. Since both surfaces are facing each other directly, the view factor \(F_{A\to B} = 1\). Now we can calculate the heat transfer rate: \(q = A \sigma (950^4 - 278.15^4) \approx 5073.87 \mathrm{W}\)

Calculate the initial rate of radiation heat transfer with aluminum foil

When the aluminum foil covers the side opening of the grill, the foil acts as a reradiating surface. We need to consider this in the calculation. The heat transfer rate can then be expressed as \(q = A \sigma F_{A\to B} (T_A^4 - T_B^4)(1+\epsilon)\), where \(\epsilon\) is the emissivity of the aluminum foil. We are not given the emissivity of the aluminum foil, but we can assume a typical value for aluminum foil, \(\epsilon \approx 0.05\). Now we can calculate the heat transfer rate with the aluminum foil cover: \(q = A \sigma (950^4 - 278.15^4)(1+0.05) \approx 5327.56 \mathrm{W}\)

Finalize the results

The initial rate of radiation heat transfer from the coal bricks to the steaks without the aluminum foil cover is approximately \(5073.87 \mathrm{W}\), and with the aluminum foil cover, it is approximately \(5327.56 \mathrm{W}\).

Key concepts

Stefan-Boltzmann Law

When we talk about radiation heat transfer, the Stefan-Boltzmann Law is a fundamental principle that comes into play. It's the cornerstone of understanding how objects emit thermal radiation.

Simply put, the Stefan-Boltzmann Law tells us that the power per unit area, also known as radiant emittance, that an object emits is directly proportional to the fourth power of the object's absolute temperature. For a black body (an idealized physical body that absorbs all incident electromagnetic radiation), the formula is represented as \(P = \sigma T^4\), where \(|P\) is the radiant emittance, \(|\sigma\) is the Stefan-Boltzmann constant \(5.67 \times 10^{-8} \text{W/m}^2\text{K}^4\), and \(|T\) is the absolute temperature in Kelvin.

Understanding this law is crucial because it allows us to calculate the rate at which an object radiates energy into the surroundings. It's particularly important in our exercise where we calculate the heat transfer from hot coal bricks to steaks using this law. The significant takeaway is that the temperature difference plays a key role; a small change in temperature can cause a large change in the heat transfer rate due to the fourth power relationship.

Blackbody Radiation

Blackbody radiation is a theoretical concept referring to an ideal object that perfectly absorbs and emits all thermal radiation. In reality, no object is a perfect blackbody, but some materials come close enough that they can be treated as such for calculations.

A blackbody is characterized by not reflecting or transmitting any radiation, which means all the heat it emits is due to its inherent temperature. This characteristic makes blackbody an excellent standard to compare other materials when we are studying thermal radiation. In the exercise, both coal bricks and steaks are approximated as blackbodies, implying that they efficiently absorb and emit radiation at the rate predicted by the Stefan-Boltzmann Law without reflection or transmission losses.

Using this assumption simplifies calculations and allows students to focus on the direct heat transfer between the objects involved, providing a clear example of how radiative energy moves from a hotter object to a cooler one.

Thermal Radiation

Thermal radiation refers to the emission of electromagnetic waves from the surface of an object, which is due to the object's temperature. It's a form of heat transfer that does not require any medium, meaning it can occur in a vacuum. This is different from conduction and convection, which require contact and a medium to transfer heat, respectively.

Everything around us emits thermal radiation, but the amount and frequency of the radiation depend on the object's temperature. Hotter objects emit more radiation, and the peak frequency of the emission shifts towards the visible spectrum as the temperature increases — this explains why objects like metal or coal bricks glow red or white when heated to high temperatures.

In the context of our problem, thermal radiation is significant because we are not considering any other form of heat transfer. While the steaks cook on the grill, they absorb thermal radiation from the hot coal bricks below. The analysis of this process involves considering both the properties of the materials involved (emissivity) and the real conditions of the environment, such as the presence of a reradiating surface like aluminum foil which alters the heat transfer scenario.