Question

On a still clear night, the sky appears to be a blackbody with an equivalent temperature of \(250 \mathrm{~K}\). What is the air temperature when a strawberry field cools to \(0^{\circ} \mathrm{C}\) and freezes if the heat transfer coefficient between the plants and air is \(6 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) because of a light breeze and the plants have an emissivity of \(0.9\) ? (a) \(14^{\circ} \mathrm{C}\) (b) \(7^{\circ} \mathrm{C}\) (c) \(3^{\circ} \mathrm{C}\) (d) \(0^{\circ} \mathrm{C}\) (e) \(-3^{\circ} \mathrm{C}\)

Short answer

Answer: (c) \(3^{\circ} \mathrm{C}\)

Step-by-step solution

Define Relevant Variables and Formulae

Let's define the variables: - Convection heat transfer: \(Q_{conv}\) - Radiation heat transfer: \(Q_{rad}\) - Heat transfer coefficient: \(h=6\; \mathrm{W/m^2\cdot K}\) - Sky temperature: \(T_{sky}=250\; \mathrm{K}\) - Emissivity of plants: \(ε=0.9\) - Stefan-Boltzmann constant: \(σ = 5.67 \times 10^{-8}\; \mathrm{W/m^2\cdot K^4}\) - Plant temperature: \(T_{plant}=0^{\circ} \mathrm{C} = 273\; \mathrm{K}\) - Air temperature: \(T_{air}\), which we need to find We will use the following equations: 1. Convection heat transfer: \(Q_{conv} = hA(T_{air} - T_{plant})\) 2. Radiation heat transfer: \(Q_{rad} = εσA(T_{plant}^4 - T_{sky}^4)\) 3. Heat balance: \(Q_{conv} = Q_{rad}\)

Equate Convection and Radiation Heat Transfers

Using the heat balance equation, equate convection and radiation heat transfers: \(hA(T_{air} - T_{plant}) = εσA(T_{plant}^4 - T_{sky}^4)\) The area A is the same on both sides and can be canceled out, so we have: \(h(T_{air} - T_{plant}) = εσ(T_{plant}^4 - T_{sky}^4)\)

Solve for Air Temperature

Now we solve the equation for \(T_{air}\): \(T_{air} - T_{plant} = \frac{εσ(T_{plant}^4 - T_{sky}^4)}{h}\) \(T_{air} = T_{plant} + \frac{εσ(T_{plant}^4 - T_{sky}^4)}{h}\) Inserting the values for \(T_{plant}\), \(T_{sky}\), \(ε\), \(σ\), and \(h\), we have: \(T_{air} = 273\; \mathrm{K} + \frac{0.9(5.67 \times 10^{-8}\; \mathrm{W/m^2\cdot K^4})((273\; \mathrm{K})^4 - (250\; \mathrm{K})^4)}{6\; \mathrm{W/m^2\cdot K}}\) \(T_{air} \approx 276\; \mathrm{K}\) Now convert back to Celsius: \(T_{air} = 276\; \mathrm{K} - 273\; \mathrm{K} = 3^{\circ} \mathrm{C}\) So, the answer is (c) \(3^{\circ} \mathrm{C}\).

Key concepts

Blackbody Radiation

Blackbody radiation refers to the idealized radiant energy a perfect black body emits. A black body is a theoretical object that absorbs all radiation it encounters, and, in turn, it radiates energy at a maximum level for any given temperature. This concept is crucial in the study of thermodynamics and quantum mechanics. It's the radiation that's emitted by a body entirely based on temperature, without considering the material properties themselves, as a perfect black body does not exist in reality.

In the context of our exercise, the sky at night is treated as a black body with an equivalent temperature of 250 K. Understanding blackbody radiation helps us calculate the energy emitted by the plants as they cool down, which is significant when considering crop temperatures during a cold night.

Convection Heat Transfer

Convection heat transfer is a mode of thermal energy transportation involving the movement of fluid (which may be a liquid or gas). This motion occurs because when a fluid is heated, it becomes less dense and rises, while the cooler fluid, being denser, sinks. As a result, there is a transfer of heat from warmer areas to cooler ones.

In our strawberry field example, there's a light breeze that affects the rate of heat transfer between the air and the plants. The heat transfer coefficient, denoted by the symbol 'h' with a value of 6 W/m²·K, represents how well the air removes heat from the surfaces of the plants. A higher heat transfer coefficient indicates a greater capability to transfer heat, which can be critical in determining the air temperature necessary to prevent the strawberry field from freezing.

Radiation Heat Transfer

Radiation heat transfer is another way energy is exchanged between surfaces and their surroundings. Unlike convection, radiation does not require a medium; it can occur in a vacuum. Heat transfer by radiation is dependent on the fourth power of the temperature (according to the Stefan-Boltzmann law), making it exponentially significant at higher temperatures.

In the textbook exercise, the plants emit thermal radiation, which is described by the Stefan-Boltzmann law and depends on factors like emissivity and the temperatures of the plants and the sky. This understanding helps us identify the rate at which the plants will lose heat during the night, which is essential to predict if and when they might freeze.

Stefan-Boltzmann Constant

The Stefan-Boltzmann constant, denoted as \(\sigma\), is a fundamental physical constant in the Stefan-Boltzmann law, which calculates the power radiated from a black body in terms of its temperature. Specifically, the equation expresses the total energy radiated per unit surface area of a black body across all wavelengths per unit time. Its value is approximately \(5.67 \times 10^{-8} W/m^2\cdot K^4\).

Incorporated into the solution for our exercise, the Stefan-Boltzmann constant is key to determining the rate at which the plants lose energy by radiation to the cool night sky, which is modeled as a black body radiator.

Emissivity

Emissivity is a measure of how effectively a surface emits thermal radiation as compared to a perfect black body. The emissivity value ranges from 0 (shiny, perfect reflectors that don't emit radiation efficiently) to 1 (perfect black bodies that emit at the maximum rate for their temperature). The emissivity of a material determines how well it radiates energy in response to its temperature.

In the exercise, the strawberry plants have an emissivity of 0.9, which means they emit radiation quite efficiently, almost like a black body. This information, in conjunction with the Stefan-Boltzmann constant and the temperature difference between the sky and the plants, allows us to calculate the radiative cooling effect on the plants. Knowing the emissivity is essential for predicting the plants' temperature on a clear night and assessing the risk of freezing.