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

(a) 1°C (b) 2°C (c) 3°C (d) 4°C Answer: (c) 3°C

Step-by-step solution

Convert temperatures to Kelvin

To work with temperatures in this exercise, we need to convert them to Kelvin. To convert from Celsius to Kelvin, we simply add 273.15: Strawberry field temperature in Kelvin: \(0^{\circ}\mathrm{C}+273.15=273.15\mathrm{K}\) Sky temperature in Kelvin: \(250\mathrm{K}\) (already in Kelvin) Now we have the temperatures in Kelvin to use in our calculations.

Calculate heat loss by radiation

We will use the Stefan-Boltzmann Law to calculate the heat loss due to radiation: \(Q_r = \varepsilon \cdot A \cdot \sigma \cdot (T_{plant}^4 - T_{sky}^4)\) Where: \(Q_r\) is the heat loss due to radiation, \(\varepsilon\) is the emissivity of the plants (0.9), \(A\) is the surface area of the plants, \(\sigma\) is the Stefan-Boltzmann constant (\(5.67 \times 10^{-8}\mathrm{W/m^2·K^4}\)), \(T_{plant}\) is the temperature of the plants in Kelvin, and \(T_{sky}\) is the temperature of the sky in Kelvin. We are looking for the air temperature, not the absolute values of heat loss, so we can write the equation without considering the surface area \(A\): \(Q_r \propto \varepsilon \cdot \sigma \cdot (T_{plant}^4 - T_{sky}^4)\) Plug in the values we know: \(Q_r \propto 0.9 \cdot 5.67 \times 10^{-8} \cdot (273.15^4 - 250^4)\)

Calculate heat loss by convection

We will use the formula for heat loss due to convection: \(Q_c = h \cdot A \cdot (T_{plant} - T_{air})\) Where: \(Q_c\) is the heat loss due to convection, \(h\) is the heat transfer coefficient (\(6\mathrm{W/m^2·K}\)), \(A\) is the surface area of the plants, and \(T_{air}\) is the air temperature in Kelvin. As in the radiation calculation, we are looking for the temperature, not the absolute values of heat loss, so we can write the equation without considering the surface area \(A\): \(Q_c \propto h \cdot (T_{plant} - T_{air})\)

Equate the heat loss by radiation and convection and solve for air temperature

When the strawberry field is in equilibrium and freezing, the heat loss due to radiation should be equal to the heat loss due to convection. Therefore, \(Q_r = Q_c\) \(\varepsilon \cdot \sigma \cdot (T_{plant}^4 - T_{sky}^4) \propto h \cdot (T_{plant} - T_{air})\) Now, we can solve for the air temperature: \(T_{air} = T_{plant} - \frac{\varepsilon \cdot \sigma \cdot (T_{plant}^4 - T_{sky}^4)}{h}\) Plug in the values: \(T_{air} = 273.15 - \frac{0.9 \cdot 5.67 \times 10^{-8} \cdot (273.15^4 - 250^4)}{6}\) \(T_{air} \approx 276.15\text{K}\)

Convert air temperature back to Celsius and select corresponding answer

Finally, we need to convert the calculated air temperature in Kelvin back to Celsius: \(T_{air} \approx 276.15\text{K} - 273.15 = 3^{\circ}\text{C}\) This corresponds to the answer (c) \(3^{\circ}\text{C}\).