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Chapter 4: Problem 166

Copper balls \(\left(\rho=8933 \mathrm{~kg} / \mathrm{m}^{3}, k=401 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, c_{p}=\right.\) \(\left.385 \mathrm{~J} / \mathrm{kg} \cdot{ }^{\circ} \mathrm{C}, \alpha=1.166 \times 10^{-4} \mathrm{~m}^{2} / \mathrm{s}\right)\) initially at \(200^{\circ} \mathrm{C}\) are allowed to cool in air at \(30^{\circ} \mathrm{C}\) for a period of 2 minutes. If the balls have a diameter of \(2 \mathrm{~cm}\) and the heat transfer coefficient is \(80 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), the center temperature of the balls at the end of cooling is (a) \(104^{\circ} \mathrm{C}\) (b) \(87^{\circ} \mathrm{C}\) (c) \(198^{\circ} \mathrm{C}\) (d) \(126^{\circ} \mathrm{C}\) (e) \(152^{\circ} \mathrm{C}\)

Short Answer

Expert verified
a) 104掳C b) 110掳C c) 95掳C d) 120掳C Answer: a) 104掳C

Step by step solution

01

Use the Newton's Law of Cooling formula

To find the temperature in the center of the balls, we will use the Newton's Law of Cooling formula for a sphere, given by: $$T(t) = T_0 + (T_i - T_0) \exp{(-\frac{h At}{\rho V c_p})}$$ where \(T(t)\) is the temperature at the end of cooling time t, \(T_0\) is the surrounding temperature, \(T_i\) is the initial temperature, \(h\) is the heat transfer coefficient, \(A\) is the surface area of the sphere, \(t\) is the cooling time, \(\rho\) is the density, \(V\) is the volume and \(c_p\) is the specific heat capacity.
02

Use the given values to calculate the temperature

We are given: \(T_i=200^{\circ} \mathrm{C}\), \(T_0=30^{\circ} \mathrm{C}\), \(h=80 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), the diameter of the copper balls \(D=2 \mathrm{~cm}\), \(\rho=8933 \mathrm{~kg} / \mathrm{m}^{3}\), \(c_{p}=385 \mathrm{~J} / \mathrm{kg} \cdot{ }^{\circ} \mathrm{C}\). First, find the values for the area \(A = \pi D^2\) and the volume \(V = \frac{4}{3} \pi (\frac{D}{2})^3\) of the copper balls.
03

Convert units and calculate the area and volume

Convert the diameter of the copper balls to meters: \(D=0.02 \mathrm{~m}\). Calculate the surface area and volume: $$A = \pi D^2 = \pi (0.02)^2 = 1.256 \times 10^{-3} \mathrm{~m}^{2}$$ $$V = \frac{4}{3} \pi (\frac{0.02}{2})^3 = 4.19 \times 10^{-6} \mathrm{~m}^{3}$$
04

Calculate the cooling time

The cooling time is given as 2 minutes, which needs to be converted to seconds: \(t=2 \times 60 = 120\) seconds.
05

Calculate the temperature at the end of cooling

Apply the values to the Newton's Law of Cooling formula: $$T(t) = 30 + (200 - 30) \exp{(-\frac{80 \times 1.256\times 10^{-3} \times 120}{8933 \times 4.19 \times 10^{-6} \times 385})}$$ $$T(t) = 30 + 170\exp{(-2.4010475)}$$ $$T(t) \approx 104^{\circ} \mathrm{C}$$So the center temperature of the balls at the end of cooling is \(104^{\circ} \mathrm{C}\) (option a).

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Key Concepts

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

Heat Transfer Coefficient
Understanding the heat transfer coefficient is crucial in thermodynamic calculations, as it provides a measure of how easily heat is transferred between materials and their surroundings. In the context of Newton's Law of Cooling, this coefficient, denoted as \( h \), quantifies the rate at which heat is lost from the surface of an object to its environment. A higher \( h \) value indicates more efficient heat loss.

For instance, in the exercise where copper balls cool in the air, the \( h \) value of \(80 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) is a property of both the copper material and the air surrounding it. The coefficient encapsulates factors like the surface roughness, the flow characteristics of the air, and other environmental conditions. It's important to understand that \( h \) is typically determined empirically and can vary based on experimental conditions.
Thermodynamics of Materials
Thermodynamics of materials refers to the study of the energy and heat associated with materials and their ability to conduct, store, and transfer heat. Key properties involved in these processes, such as specific heat capacity, thermal conductivity, and density, determine how a material responds to thermal energy.

In the exercise, copper balls exhibit specific thermodynamic characteristics. For example, copper's thermal conductivity \( k \), which is a high value of \(401 \mathrm{~W}/ \mathrm{m} \cdot \mathrm{K}\), indicates that copper is an excellent conductor of heat. This affects how fast the copper balls lose heat to their environment. The exercise leverages these thermodynamic principles to predict the cooling behavior of the balls over time.
Specific Heat Capacity
The specific heat capacity \( c_p \) of a material is the amount of heat required to raise the temperature of one kilogram of the material by one degree Celsius. It's a fundamental material property that influences thermal energy storage and changes in a material's temperature. High specific heat capacity means the material can absorb more heat before its temperature rises significantly.

In our exercise, copper's specific heat capacity of \(385 \mathrm{~J} / \mathrm{kg} \cdot{ }^{\circ} \mathrm{C}\) allows us to calculate the amount of heat energy needed for the temperature change. This is directly applied in Newton's Law of Cooling formula, illustrating its pivotal role in determining the final temperature after the cooling period.
Thermal Conductivity
Thermal conductivity is a material property that indicates how well a material conducts heat. Denoted by \( k \), it is measured in watts per meter per degree Kelvin \( \mathrm{W}/\mathrm{m} \cdot \mathrm{K} \). Materials with high thermal conductivity transfer heat quickly, whereas those with low thermal conductivity are better insulators.

As mentioned earlier, copper has a thermal conductivity of \(401 \mathrm{~W}/ \mathrm{m} \cdot \mathrm{K}\), placing it among materials with high thermal conductivity. This characteristic contributes to the rapid cooling of copper balls, as it allows heat to be efficiently distributed from the interior to the exterior of the balls, leading to a faster equalization of temperature with the surrounding air.

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Most popular questions from this chapter

Large steel plates \(1.0\)-cm in thickness are quenched from \(600^{\circ} \mathrm{C}\) to \(100^{\circ} \mathrm{C}\) by submerging them in an oil reservoir held at \(30^{\circ} \mathrm{C}\). The average heat transfer coefficient for both faces of steel plates is \(400 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Average steel properties are \(k=45 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, \rho=7800 \mathrm{~kg} / \mathrm{m}^{3}\), and \(c_{p}=470 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\). Calculate the quench time for steel plates.

It is claimed that beef can be stored for up to two years at \(-23^{\circ} \mathrm{C}\) but no more than one year at \(-12^{\circ} \mathrm{C}\). Is this claim reasonable? Explain.

A hot dog can be considered to be a \(12-\mathrm{cm}-\mathrm{long}\) cylinder whose diameter is \(2 \mathrm{~cm}\) and whose properties are \(\rho=980 \mathrm{~kg} / \mathrm{m}^{3}, c_{p}=3.9 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}, k=0.76 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), and \(\alpha=\) \(2 \times 10^{-7} \mathrm{~m}^{2} / \mathrm{s}\). A hot dog initially at \(5^{\circ} \mathrm{C}\) is dropped into boiling water at \(100^{\circ} \mathrm{C}\). The heat transfer coefficient at the surface of the hot dog is estimated to be \(600 \mathrm{~W} / \mathrm{m}^{2}\). K. If the hot dog is considered cooked when its center temperature reaches \(80^{\circ} \mathrm{C}\), determine how long it will take to cook it in the boiling water.

Hailstones are formed in high altitude clouds at \(253 \mathrm{~K}\). Consider a hailstone with diameter of \(20 \mathrm{~mm}\) and is falling through air at \(15^{\circ} \mathrm{C}\) with convection heat transfer coefficient of \(163 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Assuming the hailstone can be modeled as a sphere and has properties of ice at \(253 \mathrm{~K}\), determine the duration it takes to reach melting point at the surface of the falling hailstone. Solve this problem using analytical one-term approximation method (not the Heisler charts).

Refractory bricks are used as linings for furnaces, and they generally have low thermal conductivity to minimize heat loss through the furnace walls. Consider a thick furnace wall lining with refractory bricks \(\left(k=1.0 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\right.\) and \(\left.\alpha=5.08 \times 10^{-7} \mathrm{~m}^{2} / \mathrm{s}\right)\), where initially the wall has a uniform temperature of \(15^{\circ} \mathrm{C}\). If the wall surface is subjected to uniform heat flux of \(20 \mathrm{~kW} / \mathrm{m}^{2}\), determine the temperature at the depth of \(10 \mathrm{~cm}\) from the surface after an hour of heating time.
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