Question

A hollow spherical iron container with outer diameter \(20 \mathrm{~cm}\) and thickness \(0.2 \mathrm{~cm}\) is filled with iced water at \(0^{\circ} \mathrm{C}\). If the outer surface temperature is \(5^{\circ} \mathrm{C}\), determine the approximate rate of heat loss from the sphere, in \(\mathrm{kW}\), and the rate at which ice melts in the container. The heat of fusion of water is \(333.7 \mathrm{~kJ} / \mathrm{kg}\).

Short answer

Answer: To find the approximate rate of heat loss and the rate of ice melting, follow these steps: 1. Calculate the surface area and thickness of the sphere in meters. 2. Calculate the heat transfer through the walls of the sphere using the formula \(Q = \frac{kA\Delta T}{r_o - r_i}\). 3. Convert the heat loss rate to kW by dividing the heat transfer rate (in watts) by 1000. 4. Calculate the rate of ice melting using \(m = \frac{Q \mathrm{(in~kW)}}{333.7 \mathrm{~kJ/kg}}\). By following these steps with the given values, you can determine the approximate rate of heat loss and the rate of ice melting in the container.

Step-by-step solution

Calculate the surface area and thickness of the sphere in meters

To calculate the surface area of the sphere, we need to convert the outer diameter from centimeters to meters. Since there are 100 centimeters in a meter, we can do this by dividing by 100: Outer radius (in meters): \(r_o = \frac{20}{2} \cdot \frac{1}{100} = 0.1 \mathrm{~m}\) Now, the surface area of the sphere can be calculated using the formula: Surface area: \(A = 4 \pi r_o^2\) The thickness of the iron should also be converted to meters. Thickness (in meters): \(t = 0.2\times \frac{1}{100}= 0.002 \mathrm{~m}\) Now, we can calculate the surface area and the thickness of the sphere in meters.

Calculate the heat transfer through the walls of the sphere

In this exercise, there is a temperature difference between the inside and the outside of the sphere. Due to this difference, heat will flow through the container. To calculate the rate of heat transfer, we need to use the formula for heat flow through a sphere: \(Q = \frac{kA\Delta T}{r_o - r_i}\) where \(Q\) is the heat transfer rate, \(k\) is the thermal conductivity of the material, \(A\) is the surface area, \(\Delta T\) is the temperature difference, \(r_o\) is the outer radius of the sphere, and \(r_i\) is the inner radius of the sphere. The thermal conductivity of iron is approximately \(k = 80 \mathrm{~W/mK}\). The temperature difference is given: \(\Delta T = 5 - 0 = 5 \mathrm{~K}\). We can calculate \(r_i\) based on the known thickness and the outer radius: \(r_i = r_o - t = 0.1 - 0.002 = 0.098 \mathrm{~m}\) Now plug in the values to find the heat transfer rate: \(Q = \frac{80 \times 4 \pi (0.1)^2 \times 5}{0.1 - 0.098} \mathrm{~W}\)

Convert the heat loss rate to kW

Once we find the numerical value of the heat transfer rate (in watts), we convert it to kilowatts by dividing by 1000: \(Q \mathrm{~(in~kW)} = \frac{Q \mathrm{~(in~W)}}{1000}\)

Calculate the rate of ice melting

To calculate the rate at which the ice melts within the container, we need to relate the heat loss rate (in kilowatts) to the heat of fusion of water (in kJ/kg): Rate of ice melting (\(m\) - mass in kg/s): \(m = \frac{Q \mathrm{(in~kW)}}{333.7 \mathrm{~kJ/kg}}\) After calculating the values from steps 1 to 4, we will have the approximate rate of heat loss from the sphere in kW and the rate at which ice melts in the container.

Key concepts

Spherical Containers

When dealing with heat transfer problems involving spherical containers, it's important to understand the shape and geometry of the sphere. Spheres have the unique property of having the smallest possible surface area for a given volume. This can affect how efficiently heat is transferred through the container. In the context of a spherical container, like the iron sphere mentioned in the exercise, the diameter is a critical measurement. Here, the outer diameter is given as 20 cm, which is converted to meters for consistency in calculations. The concepts of outer radius and inner radius are crucial.

• The outer radius is half the diameter: 0.1 m in this example.
• The inner radius accounts for the material thickness, calculated by subtracting the thickness from the outer radius.

This information is necessary for calculating the sphere's surface area, which will be used in determining the rate at which heat is transferred.

Thermal Conductivity

Thermal conductivity is a material property that indicates how well a material can conduct heat. It is expressed in watts per meter per kelvin (W/mK). In this exercise, the material is iron, with a thermal conductivity value given as 80 W/mK. This property significantly affects the rate of heat transfer across a material. A higher thermal conductivity means the material will allow heat to flow more quickly through it. The rate of heat transfer due to thermal conductivity is calculated using the formula:\[ Q = \frac{kA\Delta T}{r_o - r_i} \]where:

• Q is the rate of heat transfer in watts.
• k is the thermal conductivity.
• A is the surface area, calculated previously.
• ΔT is the temperature difference.
• r_o and r_i are the outer and inner radii, respectively.

Heat of Fusion

The heat of fusion is the amount of energy per unit mass required to change a substance from solid to liquid at its melting point. For water, this value is 333.7 kJ/kg. This is an important concept when evaluating the melting process, as it provides a direct relationship between energy absorbed and mass melted. In the scenario in the exercise, we apply this concept to calculate how much ice can melt given the rate at which energy or heat is lost from the spherical container. To find the rate of ice melting:\[ m = \frac{Q \text{ (in kW)}}{333.7 \text{ kJ/kg}}\]Here, \(Q\) represents the heat transfer rate from the previous calculations, now in kilowatts. The result gives us the mass of ice converting to water per second.

Temperature Difference

Temperature difference, often denoted as ΔT, is a driving force for heat transfer. In thermal systems, heat naturally flows from regions of higher temperature to lower temperature, until thermal equilibrium is reached. In this exercise, the temperature difference (ΔT) is 5°C, calculated as the difference between the surrounding temperature and the temperature inside the spherical container (iced water, assumed to be at 0°C).Understanding ΔT is crucial as it directly affects the rate of heat transfer, according to the formula:\[ Q = \frac{kA\Delta T}{r_o - r_i} \]The greater the temperature difference, the more energy can be transferred per unit time, which can increase the rate of processes like ice melting. This principle is not only applicable to this exercise but is fundamental in any thermal analysis involving heat exchange between regions of different temperatures.