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Chapter 18: Problem 13

Why does tile feel so much colder to your feet after a bath than a bath rug? Why is this effect more striking when your feet are cold?

Short Answer

Expert verified
Answer: Tile feels colder than a rug because it has higher thermal conductivity, which means it can transfer heat more quickly. When we step on the tile with warm, wet feet, it absorbs the heat faster, making our feet lose heat more quickly and feel colder. In contrast, stepping on a rug causes slower heat transfer, keeping our feet warmer. This effect is more noticeable when our feet are already cold because the greater temperature difference between the cold feet and the tile results in even faster heat transfer, making the tile feel even colder. The slower heat transfer on a rug remains consistent, so the sensation of coldness is not as striking.

Step by step solution

01

Understand the thermal properties of materials

Different materials have different thermal properties, which includes their thermal conductivity, specific heat capacity, and thermal expansion. Thermal conductivity is a measure of how easily a material conducts heat. In this exercise, we will be primarily focusing on thermal conductivity.
02

Comparison of thermal conductivity of tile and rug

Tile is a material with higher thermal conductivity compared to a rug, which is made of fabric. In simpler terms, tile can transfer heat more quickly than a rug. When you step out of a bath with wet, warm feet, the heat from your feet transfers to the tile or rug you step onto.
03

Effect of thermal conductivity on the sensation of warmth

Since the tile has a higher thermal conductivity than a rug, it absorbs the heat from your feet faster, making your feet lose heat quickly and thus feel colder. On the other hand, when stepping on a rug, the heat transfer is slower, so your feet do not lose heat as quickly, and they feel warmer.
04

Effect of cold feet on the sensation of warmth

When your feet are cold, the temperature difference between your feet and the tile is greater than when your feet are warm. This increased temperature difference causes even faster heat transfer from your feet to the tile, making the sensation of coldness on the tile more pronounced. When stepping on a rug with cold feet, the slower heat transfer still remains, so the sensation of coldness is not as striking.

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Most popular questions from this chapter

A cryogenic storage container holds liquid helium, which boils at \(4.2 \mathrm{~K}\). Suppose a student painted the outer shell of the container black, turning it into a pseudoblackbody, and that the shell has an effective area of \(0.50 \mathrm{~m}^{2}\) and is at \(3.0 \cdot 10^{2} \mathrm{~K}\). a) Determine the rate of heat loss due to radiation. b) What is the rate at which the volume of the liquid helium in the container decreases as a result of boiling off? The latent heat of vaporization of liquid helium is \(20.9 \mathrm{~kJ} / \mathrm{kg} .\) The density of liquid helium is \(0.125 \mathrm{~kg} / \mathrm{L}\).

The radiation emitted by a blackbody at temperature \(T\) has a frequency distribution given by the Planck spectrum: $$ \epsilon_{T}(f)=\frac{2 \pi h}{c^{2}}\left(\frac{f^{3}}{e^{h f / k_{\mathrm{B}} T}-1}\right) $$ where \(\epsilon_{T}(f)\) is the energy density of the radiation per unit increment of frequency, \(v\) (for example, in watts per square meter per hertz), \(h=6.626 \cdot 10^{-34} \mathrm{~J} \mathrm{~s}\) is Planck's constant, \(k_{\mathrm{B}}=1.38 \cdot 10^{-23} \mathrm{~m}^{2} \mathrm{~kg} \mathrm{~s}^{-2} \mathrm{~K}^{-1}\) is the Boltzmann constant, and \(c\) is the speed of light in vacuum. (We'll derive this distribution in Chapter 36 as a consequence of the quantum hypothesis of light, but here it can reveal something about radiation. Remarkably, the most accurately and precisely measured example of this energy distribution in nature is the cosmic microwave background radiation.) This distribution goes to zero in the limits \(f \rightarrow 0\) and \(f \rightarrow \infty\) with a single peak in between those limits. As the temperature is increased, the energy density at each frequency value increases, and the peak shifts to a higher frequency value. a) Find the frequency corresponding to the peak of the Planck spectrum, as a function of temperature. b) Evaluate the peak frequency at temperature \(T=6.00 \cdot 10^{3} \mathrm{~K}\), approximately the temperature of the photosphere (surface) of the Sun. c) Evaluate the peak frequency at temperature \(T=2.735 \mathrm{~K}\), the temperature of the cosmic background microwave radiation. d) Evaluate the peak frequency at temperature \(T=300 . \mathrm{K}\), which is approximately the surface temperature of Earth.

Which of the following statements is (are) true? a) When a system does work, its internal energy always decreases. b) Work done on a system always decreases its internal energy. c) When a system does work on its surroundings, the sign of the work is always positive. d) Positive work done on a system is always equal to the system's gain in internal energy. e) If you push on the piston of a gas-filled cylinder, the energy of the gas in the cylinder will increase.

Knife blades are often made of hardened carbon steel. The hardening process is a heat treatment in which the blade is first heated to a temperature of \(1346^{\circ} \mathrm{F}\) and then cooled down rapidly by immersing it in a bath of water. To achieve the desired hardness, after heating to \(1346^{\circ} \mathrm{F}\), a blade needs to be brought to a temperature below \(5.00 \cdot 10^{2}{ }^{\circ} \mathrm{F}\). If the blade has a mass of \(0.500 \mathrm{~kg}\) and the water is in an open copper container of mass \(2.000 \mathrm{~kg}\) and sufficiently large volume, what is the minimum quantity of water that needs to be in the container for this hardening process to be successful? Assume the blade is not in direct mechanical (and thus thermal) contact with the container, and neglect cooling through radiation into the air. Assume no water boils but reaches \(100^{\circ} \mathrm{C} .\) The heat capacity of copper around room temperature is \(c_{\text {copper }}=386 \mathrm{~J} /(\mathrm{kg} \mathrm{K}) .\) Use the data in the table below for the heat capacity of carbon steel

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