A 15 -cm-diameter aluminum ball is to be heated from \(80^{\circ} \mathrm{C}\) to an average temperature of \(200^{\circ} \mathrm{C}\). Taking the average density and specific heat of aluminum in this temperature range to be \(\rho=2700 \mathrm{~kg} / \mathrm{m}^{3}\) and \(c_{p}=0.90 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}\), respectively, determine the amount of energy that needs to be transferred to the aluminum ball. Answer: \(515 \mathrm{~kJ}\)

Short Answer

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Specific heat of aluminum is 0.90 kJ/kg·K. Answer: 515 kJ

Step by step solution

01

Find the volume of the aluminum ball

First, we need to find the volume of the aluminum ball. The formula for the volume of a sphere is \(V = \frac{4}{3} πr^3\). In this case, the diameter of the ball is 15 cm, so the radius is 7.5 cm, or 0.075 m. Using this radius, we can calculate the volume: $$ V = \frac{4}{3} π (0.075 \mathrm{m})^3 $$
02

Calculate the mass of the aluminum ball

Now, we will calculate the mass of the ball, using the formula \(m = ρV\), where \(ρ\) is the density and \(V\) is the volume. We are given the density, 2700 kg/m³. Using the volume we calculated in Step 1, we can find the mass: $$ m = (2700 \mathrm{~kg/m^3}) (V) $$
03

Calculate the change in temperature

We are given that the initial temperature of the aluminum ball is 80°C, and we want to heat it to an average temperature of 200°C. To find the change in temperature, subtract the initial temperature from the final temperature: $$ ΔT = 200^{\circ} \mathrm{C} - 80^{\circ} \mathrm{C} $$
04

Calculate the amount of energy needed

Finally, we can calculate the amount of energy needed to heat the aluminum ball to the desired temperature. We will use the formula \(Q = mcΔT\), where \(c\) is the specific heat. We are given the specific heat, 0.90 kJ/kg·K (which is equal to 900 J/kg·K). Plugging in the values for mass, specific heat, and change in temperature, we find the energy needed: $$ Q = m(900 \mathrm{~J/kg\cdot K}) (ΔT) $$ Use the values calculated in previous steps, and the result will be: $$ Q = 515 \mathrm{~kJ} $$ The amount of energy needed to heat the aluminum ball from 80°C to 200°C is 515 kJ.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Specific Heat
Specific heat, symbolized as c, is a material property that stands for the amount of heat needed to raise the temperature of one kilogram of a substance by one degree Celsius (or Kelvin). In other words, it is a measure of how much thermal energy a substance can soak up before its temperature changes significantly.

The specific heat of a substance is a critical aspect in heat transfer calculations because it reflects how substances absorb and store heat. For instance, in the given exercise, the specific heat of aluminum was given as 0.90 kJ/kg\(\bullet\)K. This relatively low specific heat means that aluminum heats up quickly compared to substances with a higher specific heat. The ability of a substance to resist temperature changes also has practical applications, such as in constructing buildings or manufacturing cookware.
Thermal Energy Transfer
Thermal energy transfer involves moving heat from one object or substance to another. The method of transfer can occur in three primary ways: conduction, convection, and radiation. Conduction happens through direct contact, convection is through fluid movement (like air or water), and radiation transfers energy through electromagnetic waves.

In the textbook exercise, the quantity of thermal energy that must be transferred to the aluminum ball to heat it is what we are solving for, denoted as Q. The formula used in the exercise, Q = mc\(\Delta\)T, succinctly expresses thermal energy transfer in cases of conduction or convection where no phase change occurs. It states that the amount of energy transferred (Q) is the product of the mass (m), specific heat (c), and change in temperature (\(\Delta\)T).
Temperature Change
Temperature change is a concept that measures the difference in temperature as a substance absorbs or releases heat. It is denoted as \(\Delta\)T in heat transfer equations, calculating the shift in temperature from an initial state T1 to a final state T2, described as \(\Delta\)T = T2 - T1.

This variable is essential in understanding how much energy is required to achieve a specific temperature increase, as seen in the exercise where the aluminum ball’s temperature rises from 80°C to 200°C. The broader the temperature change, the more energy is typically needed, provided the mass and specific heat remain constant. This relationship is crucial in various applications, from climate control in buildings to managing thermal conditions in industrial processes.

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Most popular questions from this chapter

An aluminum pan whose thermal conductivity is \(237 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) has a flat bottom with diameter \(15 \mathrm{~cm}\) and thickness \(0.4 \mathrm{~cm}\). Heat is transferred steadily to boiling water in the pan through its bottom at a rate of \(1400 \mathrm{~W}\). If the inner surface of the bottom of the pan is at \(105^{\circ} \mathrm{C}\), determine the temperature of the outer surface of the bottom of the pan.

Consider a flat-plate solar collector placed horizontally on the flat roof of a house. The collector is \(5 \mathrm{ft}\) wide and \(15 \mathrm{ft}\) long, and the average temperature of the exposed surface of the collector is \(100^{\circ} \mathrm{F}\). The emissivity of the exposed surface of the collector is \(0.9\). Determine the rate of heat loss from the collector by convection and radiation during a calm day when the ambient air temperature is \(70^{\circ} \mathrm{F}\) and the effective sky temperature for radiation exchange is \(50^{\circ} \mathrm{F}\). Take the convection heat transfer coefficient on the exposed surface to be \(2.5 \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft}^{2} \cdot{ }^{\circ} \mathrm{F}\).

The rate of heat loss through a unit surface area of a window per unit temperature difference between the indoors and the outdoors is called the \(U\)-factor. The value of the \(U\)-factor ranges from about \(1.25 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) (or \(0.22 \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft}^{2}{ }^{\circ} \mathrm{F}\) ) for low-e coated, argon-filled, quadruple-pane windows to \(6.25 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) (or \(1.1 \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft}^{2}{ }^{\circ} \mathrm{F}\) ) for a single-pane window with aluminum frames. Determine the range for the rate of heat loss through a \(1.2-\mathrm{m} \times 1.8-\mathrm{m}\) window of a house that is maintained at \(20^{\circ} \mathrm{C}\) when the outdoor air temperature is \(-8^{\circ} \mathrm{C}\).

In a power plant, pipes transporting superheated vapor are very common. Superheated vapor is flowing at a rate of \(0.3 \mathrm{~kg} / \mathrm{s}\) inside a pipe with \(5 \mathrm{~cm}\) in diameter and \(10 \mathrm{~m}\) in length. The pipe is located in a power plant at \(20^{\circ} \mathrm{C}\), and has a uniform pipe surface temperature of \(100^{\circ} \mathrm{C}\). If the temperature drop between the inlet and exit of the pipe is \(30^{\circ} \mathrm{C}\), and the specific heat of the vapor is \(2190 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\), determine the heat transfer coefficient as a result of convection between the pipe surface and the surrounding.

The inner and outer surfaces of a \(0.5-\mathrm{cm}\) thick \(2-\mathrm{m} \times 2-\mathrm{m}\) window glass in winter are \(10^{\circ} \mathrm{C}\) and \(3^{\circ} \mathrm{C}\), respectively. If the thermal conductivity of the glass is \(0.78 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), determine the amount of heat loss through the glass over a period of \(5 \mathrm{~h}\). What would your answer be if the glass were \(1 \mathrm{~cm}\) thick?

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