Problem 108 - University Physics with Modern Physics

Question

A metal sphere with radius 3.20 cm is suspended in a large metal box with interior walls that are maintained at 30.0\(^\circ\)C. A small electric heater is embedded in the sphere. Heat energy must be supplied to the sphere at the rate of 0.660 J/s to maintain the sphere at a constant temperature of 41.0\(^\circ\)C. (a) What is the emissivity of the metal sphere? (b) What power input to the sphere is required to maintain it at 82.0\(^\circ\)C? What is the ratio of the power required for 82.0\(^\circ\)C to the power required for 41.0\(^\circ\)C? How does this ratio compare with 2\(^4\)? Explain.

Step-by-Step Solution

Answer

(a) \(\varepsilon \approx 0.013\); (b) \(P_{82} \approx 2.90 \text{ W}\); Ratio: \(\approx 4.39\), less than \(2^4 = 16\).

Step 1: Understand Problem Requirements

The problem involves determining the emissivity of a metal sphere using given parameters such as the temperature difference between the sphere and surroundings, and the rate of heat supply. Additionally, we need to find the power needed to maintain the sphere at a higher temperature and compare power ratios.

Step 2: Relevant Formula for Emissivity and Power

The Stefan-Boltzmann law relates power radiated by a surface to temperature and emissivity: \[ P = \varepsilon \sigma A (T_{sphere}^4 - T_{walls}^4) \]where \(P\) is the power (in watts), \(\varepsilon\) is emissivity, \(\sigma\) is the Stefan-Boltzmann constant \((5.67 \times 10^{-8} \text{ W/m}^2\cdot\text{K}^4)\), \(A\) is the surface area, and \(T\) represents temperatures in Kelvin.

Step 3: Convert All Temperatures to Kelvin

Convert the temperatures to Kelvin:- \(T_{sphere} = 41.0^\circ\text{C} = 41 + 273.15 = 314.15\text{ K}\)- \(T_{walls} = 30.0^\circ\text{C} = 30 + 273.15 = 303.15\text{ K}\)

Step 4: Calculate Surface Area of Sphere

The surface area \(A\) of a sphere is given by:\[ A = 4\pi r^2 \]where \(r = 3.20\text{ cm} = 0.032\text{ m}\), so,\[ A = 4\pi (0.032)^2 \approx 0.0129\text{ m}^2 \]

Step 5: Calculate Emissivity

Substitute known values into the Stefan-Boltzmann law to find emissivity \(\varepsilon\):\[ 0.660 = \varepsilon (5.67 \times 10^{-8}) (0.0129) (314.15^4 - 303.15^4) \]Solving for \(\varepsilon\), we find:\[ \varepsilon \approx 0.013 \]

Step 6: Calculate Power for 82.0°C

Repeat the calculation with \(T_{sphere} = 82.0^\circ\text{C} = 355.15\text{ K}\):\[ P_{82} = \varepsilon (5.67 \times 10^{-8}) (0.0129) (355.15^4 - 303.15^4) \]Substituting \(\varepsilon = 0.013\) and solving, we find:\[ P_{82} \approx 2.90 \text{ W} \]

Step 7: Calculate Power Ratio and Compare with 2^4

Find the ratio of powers:\[ \frac{P_{82}}{P_{41}} = \frac{2.90}{0.660} \approx 4.39 \]Compare this ratio with \(2^4 = 16\). Since \(4.39\) is less than \(16\), the relationship is mainly due to non-linear increase in radiative power with temperature.

Key Concepts

Stefan-Boltzmann Law

The Stefan-Boltzmann Law provides a fundamental principle for understanding how objects radiate heat. It states that the power radiated by an object is proportional to the fourth power of its absolute temperature. This equation is given by:\[ P = \varepsilon \sigma A (T_{sphere}^4 - T_{walls}^4) \] - **Where:** - \(P\) is the radiant power in watts - \(\varepsilon\) is the emissivity of the surface, a measure of how effectively an object radiates heat - \(\sigma\) is the Stefan-Boltzmann constant, approximately \(5.67 \times 10^{-8} \text{ W/m}^2\cdot\text{K}^4\) - \(A\) is the surface area of the radiating object - \(T\) represents temperature in KelvinThis law is crucial in thermal physics because it relates the emissivity and the temperature to the power, enabling us to calculate parameters like the heat supplied or required by a system. It’s especially important for objects not perfectly black, meaning not all objects absorb and emit all radiation equally, like the metal sphere in our exercise. Calculating \(\varepsilon\) using this law allows us to understand how efficiently the sphere emits thermal radiation.

Thermal Radiation

Thermal radiation is the emission of electromagnetic waves from all substances that have a temperature above absolute zero. This process explains how heat energy is transferred through space even in the absence of contact, which is crucial in fields like astrophysics, climate science, and everyday heating.Here are some key points about thermal radiation:- **Emanation from Surface:** - All objects emit thermal radiation as long as they have a temperature. The energy radiated depends on the temperature and properties of the object's surface. - In the exercise, the metal sphere emits radiation which we need to account when balancing the heat provided by the heater.- **Influences by Emissivity:** Emissivity \((\varepsilon)\) is a measure of a surface’s efficiency at emitting energy as thermal radiation. It ranges from 0 (perfect reflector) to 1 (perfect emitter or blackbody). The exercise focuses on finding this parameter, showing it's essential to predict how well a real object like the sphere radiates energy compared to an ideal blackbody.- **Dependence on Temperature:** - The amount and type of thermal radiation depend significantly on the temperature. As objects heat up, their radiation increases dramatically according to the Stefan-Boltzmann law, highlighting why understanding the emissivity and the environment temperature difference is necessary for our exercise.These principles help explain the physics behind maintaining the temperature of the sphere, showing how understanding thermal radiation and its calculation is vital in defining the thermal landscape of an environment.

Power Calculation

Calculating the power required to heat an object involves determining how much energy is needed per second to achieve a desired temperature. In our exercise, the concept of power calculation is applied to the metal sphere to ensure it maintains a specific temperature despite losing heat through radiation.Understanding power calculation involves several steps:- **Determine Current and Desired States:** - Knowing the initial and target temperatures is crucial. For instance, we first calculate power for the sphere at 41°C and later adjust it for 82°C. - This involves recalculating the temperature difference with the surrounding environment and substituting back into the Stefan-Boltzmann equation. - **Substitute into Derived Formula:** - Using \(P = \varepsilon \sigma A (T_{sphere}^4 - T_{walls}^4)\), substitute your known values to calculate \(P\). - Our steps show how emissivity and area are constants for the sphere, and how power changes with temperature.- **Compare Results:** - After finding the power for both temperatures, the next step is to compare these values to understand trends and verify if calculations align with expectations such as potential experimental results or theoretical figures like \(2^4\). - In the exercise, this step helps us understand the significance of nonlinear increases in power requirements with temperature changes. By following these elements, we can accurately determine the required power input to maintain an object’s temperature, a critical skill in thermal engineering and sciences concerning energy management.