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Chapter 4: Problem 11

Consider a sphere and a cylinder of equal volume made of copper. Both the sphere and the cylinder are initially at the same temperature and are exposed to convection in the same environment. Which do you think will cool faster, the cylinder or the sphere? Why?

Short Answer

Expert verified
Answer: The choice of which object cools faster depends on the specific dimensions of the cylinder and sphere under consideration. By utilizing the ratio of surface areas between the sphere and the cylinder, we could determine which object cools faster for particular dimensions of the two shapes. The object with a larger surface area will cool at a faster rate.

Step by step solution

01

Identifying the given information

In this problem, we are given that both the sphere and the cylinder have the same volume and are made of the same material, copper. They are also initially at the same temperature and experience convection in the same environment. The rate of cooling is affected by their surface area.
02

Relate the volume of the sphere and the cylinder

Denote the radius of the sphere as r and the radius and height of the cylinder as R and H respectively. Since the volumes of the sphere and the cylinder are equal, their relationship is given by: (volume of sphere) = (volume of cylinder) \[\frac{4}{3}\pi r^3 = \pi R^2 H\]
03

Calculate the surface area of the sphere and the cylinder

Now, we need to determine the surface area of the sphere and the cylinder. The surface area of the sphere, A_s, is given by: \[A_s = 4 \pi r^2\] The surface area of the cylinder, A_c, is given by: \[A_c = 2\pi R (R + H)\]
04

Find the ratio of surface areas

To determine which object cools faster, we will compare their surface areas. Divide the surface area of the sphere by the surface area of the cylinder: \[\frac{A_s}{A_c} = \frac{4 \pi r^2}{2 \pi R (R + H)}\] Since the surface area of both shapes consists of the same material and environment, the pi terms cancel out: \[\frac{A_s}{A_c} = \frac{2 r^2}{R (R + H)}\]
05

Identify which object has a larger surface area

Using the relationship between the volumes of the sphere and the cylinder, determine the ratio of their radii as follows: \[\frac{r^3}{R^2 H} = \frac{4}{3}\] Combine this ratio with the surface area ratio: \[\frac{A_s}{A_c} = \frac{6R}{R + H}\] For the sphere to have a larger surface area and hence cool faster, the ratio should be greater than 1. If the ratio is less than 1, the cylinder will have a larger surface area and cool faster. In other words, - If \[\frac{6R}{R + H} > 1\], the sphere cools faster. - If \[\frac{6R}{R + H} < 1\], the cylinder cools faster.
06

Conclusion

The choice of which object cools faster depends on the specific dimensions of the cylinder and sphere under consideration. However, the relationship between the objects' surface area and the rate of heat loss indicates that the one with a larger surface area will cool at a faster rate. Utilizing the ratio of surface areas between the sphere and the cylinder, we could determine which object cools faster for particular dimensions of the two shapes.

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Key Concepts

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

Thermal Conductivity of Copper
Thermal conductivity is a material's ability to conduct heat, and copper is known for its excellent thermal conductivity. It's important for students to understand that when a solid like copper loses heat, this process is largely governed by how easily heat can move through the material. Copper, having high thermal conductivity, means that heat is quickly distributed throughout the material and then dissipated to the surroundings.

In the context of our problem comparing the cooling rates of a copper sphere and cylinder, this property ensures that the initial uniform temperature within both shapes will promote an even rate of heat transfer to the surface, where heat loss occurs through convection. Thus, regardless of shape, the intrinsic property of copper to rapidly conduct heat plays a crucial role in the cooling process.
Surface Area to Volume Ratio

Understanding the Ratio

The surface area to volume ratio is pivotal in determining the rate of cooling for objects. Essentially, this ratio indicates how much surface area is available for heat transfer in relation to the object's size.

Cooling Implications

An object with a high surface area to volume ratio will generally cool faster because there is more surface through which heat can be expelled, per unit of volume. For example, when students examine smaller objects, or objects with protrusions, they find an increased surface area in relation to volume, which leads to a faster rate of cooling in comparison to objects with lower ratios.
Sphere and Cylinder Heat Dissipation
The dissipation of heat from objects like spheres and cylinders is a visual representation of the concepts of surface area, volume, and thermal conductivity at play. Specific to our problem, while the copper material of the sphere and cylinder ensures fast heat transfer through their volume, the rate of heat dissipation is not only dependent on the material but also the shape.

Typically, a sphere presents a lower surface area to volume ratio compared to a cylinder with equivalent volume, suggesting that a sphere would generally dissipate heat more slowly. However, this relationship can vary with the dimensions of the cylinder鈥攖hat is, the ratio of its height to its radius.
Rate of Cooling in Solids
The rate of cooling in solids links directly back to the principles of thermal conductivity and surface area to volume ratio. It's a measure of how quickly an object's temperature decreases over time. Newton's Law of Cooling indicates that the rate of cooling is proportional to the temperature difference between the object and its environment, assuming that this difference is relatively small.

Furthermore, solids like our copper shapes will also cool at rates influenced by their specific surface areas. Larger surface area allows for more heat to be transferred to the surrounding environment, hence a faster cooling rate. To help students visualize this, one might consider a hot plate of copper cooling faster than a block of the same material, due to the plate's greater surface area in contact with the air.

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Most popular questions from this chapter

How can we use the transient temperature charts when the surface temperature of the geometry is specified instead of the temperature of the surrounding medium and the convection heat transfer coefficient?

Why are the transient temperature charts prepared using nondimensionalized quantities such as the Biot and Fourier numbers instead of the actual variables such as thermal conductivity and time?

A 6-cm-diameter 13-cm-high canned drink ( \(\rho=\) \(\left.977 \mathrm{~kg} / \mathrm{m}^{3}, k=0.607 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, c_{p}=4180 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\) initially at \(25^{\circ} \mathrm{C}\) is to be cooled to \(5^{\circ} \mathrm{C}\) by dropping it into iced water at \(0^{\circ} \mathrm{C}\). Total surface area and volume of the drink are \(A_{s}=\) \(301.6 \mathrm{~cm}^{2}\) and \(V=367.6 \mathrm{~cm}^{3}\). If the heat transfer coefficient is \(120 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), determine how long it will take for the drink to \(\operatorname{cool}\) to \(5^{\circ} \mathrm{C}\). Assume the can is agitated in water and thus the temperature of the drink changes uniformly with time. (a) \(1.5 \mathrm{~min}\) (b) \(8.7 \mathrm{~min}\) (c) \(11.1 \mathrm{~min}\) (d) \(26.6 \mathrm{~min}\) (e) \(6.7 \mathrm{~min}\)

What are the factors that affect the quality of frozen fish?

Consider a hot semi-infinite solid at an initial temperature of \(T_{i}\) that is exposed to convection to a cooler medium at a constant temperature of \(T_{\infty}\), with a heat transfer coefficient of \(h\). Explain how you can determine the total amount of heat transfer from the solid up to a specified time \(t_{o}\).
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