Problem 58 - College Physics

Question

A box-shaped wood stove has dimensions of 0.75 \(\mathrm{m} \times\) \(1.2 \mathrm{m} \times 0.40 \mathrm{m},\) an emissivity of \(0.85,\) and a surface temperature of \(205^{\circ} \mathrm{C}\) . Calculate its rate of radiation into the surrounding space.

Step-by-Step Solution

Answer

The rate of radiation from the stove is approximately \(6849.65 \mathrm{W}\).

Step 1: Convert Surface Temperature

The surface temperature of the stove is given as \(205^{\circ} \mathrm{C}\). To use it in the formula for radiation, we must convert it to Kelvin. The conversion formula is:\[ T(\mathrm{K}) = T(^{\circ} \mathrm{C}) + 273.15 \]So, \(T = 205 + 273.15 = 478.15 \mathrm{K}\).

Step 2: Calculate Surface Area

The stove's dimensions are \(0.75 \mathrm{m} \times 1.2 \mathrm{m} \times 0.40 \mathrm{m}\). To find its total surface area, use the formula for the surface area of a rectangular box:\[ A = 2(lw + lh + wh) \]where \(l = 0.75\, \mathrm{m},\, w = 1.2\, \mathrm{m},\, h = 0.40\, \mathrm{m}\).Thus, \(A = 2((0.75 \times 1.2) + (0.75 \times 0.4) + (1.2 \times 0.4)) = 2(0.9 + 0.3 + 0.48) = 2 \times 1.68 = 3.36 \mathrm{m}^2\).

Step 3: Use Stefan-Boltzmann Law to Calculate Radiated Power

The rate of radiation emitted by the stove can be calculated using the Stefan-Boltzmann Law:\[ P = \varepsilon \sigma A T^4 \]where:- \(\varepsilon = 0.85\) (emissivity),- \(\sigma = 5.67 \times 10^{-8} \mathrm{W}\, \mathrm{m}^{-2}\, \mathrm{K}^{-4}\) (Stefan-Boltzmann constant),- \(A = 3.36 \mathrm{m}^2\) (surface area),- \(T = 478.15 \mathrm{K}\) (temperature in Kelvin).Substituting these values, we find:\[ P = 0.85 \times 5.67 \times 10^{-8} \times 3.36 \times (478.15)^4 \approx 6849.65 \mathrm{W}\].

Key Concepts

Emissivity

Emissivity is a fundamental concept in understanding how objects emit thermal radiation. It measures the efficiency with which a surface emits energy as thermal radiation compared to a perfect black body at the same temperature. A black body is an idealized object that absorbs and radiates all energy it receives. Emissivity values range between 0 and 1.

- **Understanding emissivity:** A value of 1 indicates a perfect emitter (like a black body), while 0 suggests no thermal radiation emitted. Real-world materials have emissivity values between these two extremes.
- **Importance in the Stefan-Boltzmann Law:** Emissivity directly impacts the rate at which an object radiates energy. High emissivity materials, like the wood stove with 0.85 emissivity, are efficient at emitting thermal radiation.

In practical terms, emissivity helps us predict and calculate the energy loss of objects in various environments. Knowing the emissivity of an object lets us estimate how much heat is radiated naturally, a crucial factor in calculating heat loss in heating appliances, industrial processes, and even in measuring our planet's energy balance.

Surface Area Calculation

Calculating the surface area of an object, especially for irregular shapes, is important for tasks like assessing heat radiation or conduction. For simple geometric shapes like our box-shaped wood stove, the calculation is straightforward using the formula for a rectangle.

- **Formula and Process:** The formula for a rectangular prism's surface area is: \[ A = 2(lw + lh + wh) \ \] Where \(l\), \(w\), and \(h\) are the length, width, and height, respectively.

For our wood stove, with dimensions of 0.75 m × 1.2 m × 0.40 m:

• Top and bottom surface: \(2 imes (l imes w) = 2 imes (0.75 imes 1.2) = 1.8 \, \text{m}^2\)
• Front and back surface: \(2 imes (l imes h) = 2 imes (0.75 imes 0.4) = 0.6 \, \text{m}^2\)
• Sides: \(2 imes (w imes h) = 2 imes (1.2 imes 0.4) = 0.96 \, \text{m}^2\)

Adding these results gives the total surface area of 3.36 \(\text{m}^2\).

Understanding how to calculate surface area helps in determining the heat exchange capacity of objects, affecting everything from insulation efficiency to the material's design used in constructions.

Temperature Conversion

Temperature conversion is essential in various scientific calculations, including those involving thermodynamics and radiative heat transfer. Converting temperatures between Celsius and Kelvin or Fahrenheit is often necessary for applying different physics formulas correctly.

- **From Celsius to Kelvin:** The Kelvin scale is the base unit for temperature in the International System of Units (SI). It starts at absolute zero, where all thermal motion ceases. To convert a temperature from Celsius to Kelvin, simply add 273.15: \[ T(\text{K}) = T(^{\circ} \text{C}) + 273.15 \ \] This formula is handy for theoretical and practical physics problems.

For the exercise, converting the stove's surface temperature from 205°C to Kelvin was done by adding 273.15, resulting in 478.15 K. Such conversions are critical for formulas like the Stefan-Boltzmann Law, which requires temperature input in Kelvin for calculating the radiant energy from an object accurately.

By mastering temperature conversions, students gain the flexibility to tackle various scientific problems, enabling clearer communication of thermal measurements and ensuring all calculations align with standard scientific practices.