Question

Consider a black spherical ball, with a diameter of \(25 \mathrm{~cm}\), is being suspended in air. Determine the surface temperature of the ball that should be maintained in order to heat \(10 \mathrm{~kg}\) of air from 20 to \(30^{\circ} \mathrm{C}\) in the duration of 5 minutes.

Short answer

Answer: The surface temperature of the black spherical ball should be approximately 106.25°C.

Step-by-step solution

Calculate the required energy

First, we need to find out how much energy is required to heat the air. To do this, we can use the equation: $$Q = mcΔT$$ Where: - \(Q\) is the heat energy required (Joules) - \(m\) is the mass of the air (kg) - \(c\) is the specific heat capacity of the air (\(\frac{J}{kgK}\)) - \(ΔT\) is the temperature difference (K) Here, \(m = 10\mathrm{~kg}\), and \(ΔT = 30 - 20 = 10\mathrm{~K}\). The specific heat capacity of the air is approximately \(c = 1000\frac{J}{kgK}\). So, the heat energy required is: $$Q = 10\mathrm{~kg} \times 1000\frac{J}{kgK} \times 10\mathrm{~K} = 100,000\mathrm{~J}$$

Calculate the heat emission rate

Now, we must find how fast the ball has to emit heat to transfer the required energy within the given time. The given time is 5 minutes, or \(t = 300\mathrm{~seconds}\). The heat emission rate \(P\) can be found using: $$P = \frac{Q}{t}$$ $$P = \frac{100,000\mathrm{~J}}{300\mathrm{~seconds}} = 333.33\mathrm{~\frac{W}{s}}$$

Use Stefan-Boltzmann Law and find the surface temperature

The Stefan-Boltzmann Law states that the power emitted by a black body (the ball in our case) is given by: $$P = A \sigma T^4$$ Where: - \(P\) is the power emitted (W) - \(A\) is the surface area of the object (m²) - \(σ\) is the Stefan-Boltzmann constant (\(5.67 \times 10^{-8} \frac{W}{m^2K^4}\)) - \(T\) is the temperature of the object (K) In this problem, we need to find \(T\). First, we must find the surface area of the sphere, where the diameter is given to be \(25\mathrm{~cm}\) (or \(0.25\mathrm{~m}\)). The surface area of a sphere is given by: $$A = 4πR^2$$ Where \(R\) is the radius of the sphere. Here, \(R = 0.125\mathrm{~m}\) and thus: $$A = 4π(0.125\mathrm{~m})^2 = 0.196\mathrm{~m^2}$$ Now, we can rearrange the Stefan-Boltzmann Law to find temperature: $$T^4 = \frac{P}{A\sigma}$$ $$T^4 = \frac{333.33\mathrm{~\frac{W}{s}}}{0.196\mathrm{~m^2} \times 5.67 \times 10^{-8} \frac{W}{m^2K^4}} = 186249228.72$$ $$T = \sqrt[4]{186249228.72} = 379.40\mathrm{~K}$$

Convert the temperature to Celsius for the final answer

Finally, we need to convert the surface temperature from Kelvin to Celsius using the following equation: $$T_{C} = T_{k} - 273.15$$ $$T_{C} = 379.40\mathrm{~K} - 273.15 = 106.25^{\circ}\mathrm{C}$$ The surface temperature of the black spherical ball that should be maintained to heat the air from \(20^{\circ} \mathrm{C}\) to \(30^{\circ} \mathrm{C}\) in 5 minutes is approximately \(106.25^{\circ} \mathrm{C}\).

Key concepts

Specific Heat Capacity

Specific heat capacity is an essential concept in heat transfer calculations as it defines the amount of heat energy required to raise the temperature of a certain amount of a substance by one degree Celsius or one Kelvin. It's denoted with the symbol 'c' and its unit is joules per kilogram per Kelvin \(\frac{J}{kgK}\).

In the exercise, we used the specific heat capacity of air, which is approximately \(1000\frac{J}{kgK}\). This means that to increase the temperature of one kilogram of air by one Kelvin, one thousand joules of energy are needed. The cumulative energy, \(Q\), employed to warm up the air (with mass 'm' and experiencing a temperature change \(\Delta T\)) can be calculated using the formula \(Q = mc\Delta T\).
This formula ensures a straightforward computation of the heat energy needed to reach a specific temperature rise given the air's mass and specific heat capacity, resulting in \(Q = 10\ kg \times 1000\frac{J}{kgK} \times 10\ K = 100,000\ J\). Understanding the specific heat capacity is critical as it underscores the energy efficiency or inefficiency in the heating process of various materials.

Stefan-Boltzmann Law

The Stefan-Boltzmann Law provides insight into the fundamental principles of thermodynamics and radiation heat transfer. This law indicates that the power radiated per unit area of a black body is directly proportional to the fourth power of the body’s absolute temperature.

The law is often represented by the equation \(P = A \sigma T^4\), where \(P\) refers to the radiated power, \(A\) denotes the surface area of the radiating body, \(\sigma\) is the Stefan-Boltzmann constant \(5.67 \times 10^{-8} \frac{W}{m^2K^4}\), and \(T\) stands for the absolute temperature in Kelvin.
In our exercise involving a spherical object, the law allowed us to deduce the required temperature for heating air by first solving for the surface area and subsequently the power emitted. By rearranging the variables and solving for temperature, we demonstrate how crucial the precise calculation of radiated power is for controlling temperatures in practical applications. The accuracy of this law is paramount for predicting the heat exchange between objects and their environments in complex systems.

Spherical Object Heating

When calculating the heating effect of a spherical object, understanding how surface area and temperature interplay is fundamental. A sphere's surface area impacts the amounts of heat it can exchange with its surroundings because a greater area provides more 'space' for the heat transfer.

To calculate the spherical surface area, we apply the formula \(A = 4\pi R^2\), where \(A\) is the area and \(R\) is the sphere's radius. In our exercise, the black spherical ball has a diameter of 25 cm, giving it a radius of 12.5 cm or 0.125 m. The resulting surface area, which is important to the Stefan-Boltzmann Law, turns out to be \(0.196 m^2\).
With this step towards calculating the ball's emission rate, we move closer to defining how much of its surface temperature needs to be maintained to heat up air within a specific time frame. Such careful attention to heating processes and surface area in the context of spherical objects is pivotal across various fields, from astrophysical observations to industrial processes like the manufacture of heating elements.