/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 50 A copper teakettle has a circula... [FREE SOLUTION] | 91影视

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Chapter 11: Problem 50

A copper teakettle has a circular bottom \(30.0 \mathrm{~cm}\) in diameter that has a uniform thickness of \(2.50 \mathrm{~mm}\). It sits on a burner whose temperature is \(150^{\circ} \mathrm{C}\). (a) If the teakettle is full of boiling water, what is the rate of heat conduction through its bottom? (b) Assuming that the heat from the burner is the only heat input, how much water is boiled away in \(5.0 \mathrm{~min} ?\) Is your answer unreasonably large? If yes, explain why.

Short Answer

Expert verified
(a) The rate of heat conduction is 568,414 W. (b) 75.7 kg of water boils away, which is unrealistically large due to ignored heat losses.

Step by step solution

01

Calculate the Area of the Bottom

The diameter of the teakettle's bottom is given as 30.0 cm, so the radius is half of that: \( r = \frac{30.0}{2} = 15.0 \) cm = 0.15 m. Use the formula for the area of a circle, \( A = \pi r^2 \), to find the area:\[ A = \pi (0.15)^2 = 0.0707 \text{ m}^2 \]
02

Find the Temperature Difference

The problem states that the burner temperature is \(150^{\circ}\text{C}\) and the water is boiling at \(100^{\circ}\text{C}\). The temperature difference \( \Delta T \) is:\[ \Delta T = 150 - 100 = 50^{\circ}\text{C} \]
03

Use Fourier's Law of Heat Conduction

Fourier's Law states that \( Q = \frac{kA\Delta T}{d} \), where \( k \) is the thermal conductivity of copper \( (k = 401 \text{ W/m/K}) \), \( A \) is the area, \( \Delta T \) is the temperature difference, and \( d \) is the thickness (converted to meters): \( d = 2.50 \text{ mm} = 0.0025 \text{ m} \). Substitute these values into the equation:\[ Q = \frac{401 \times 0.0707 \times 50}{0.0025} \approx 568414 \text{ W} \]
04

Calculate the Mass of Water Boiled Away

The latent heat of vaporization of water is \( 2260 \text{ J/g} \). The energy conducted through the bottom per second is \( 568414 \text{ W} \). Over \(5.0 \text{ minutes} = 300 \text{ seconds} \), the total energy is:\[ E = 568414 \times 300 \approx 1.71 \times 10^8 \text{ J} \]Convert this energy to mass:\[ m = \frac{E}{L_v} = \frac{1.71 \times 10^8}{2260} \approx 75664 \text{ g} = 75.7 \text{ kg} \]
05

Determine Reasonableness of the Answer

The calculated mass of 75.7 kg is much larger than what can be reasonably expected. The conclusion is that this rate of heat conduction through the kettle's bottom is unrealistically high, possibly due to ignoring losses to the surroundings.

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Key Concepts

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

Thermal Conductivity
Thermal conductivity is a material's innate ability to conduct heat. It is represented by the symbol \( k \). High thermal conductivity means a material can efficiently transfer heat. Copper, with a thermal conductivity of \( 401 \text{ W/m/K} \), is an excellent conductor. This property makes copper ideal for cookware bottoms, like the teakettle in the exercise. Thermal conduction involves energy transfer due to temperature differences. Think of \( k \) as a measure of how well the bottom of the teakettle can carry heat from the burner to the water. The design of such a kettle should maximize heat transfer, ensuring quick boiling times.
Fourier's Law
Fourier's Law of Heat Conduction provides a mathematical framework to understand heat flow. The law states that the rate of heat conduction \( Q \) through a material is proportional to the temperature difference \( \Delta T \) and the area \( A \) through which heat flows, inversely proportional to the material's thickness \( d \). The formula is: \[ Q = \frac{kA\Delta T}{d} \]Where:
  • \( k \) is thermal conductivity
  • \( A \) is the area
  • \( \Delta T \) is the temperature difference
  • \( d \) is the thickness
In our exercise, this law helps determine the heat transfer rate from the burner's surface to boiling water inside the kettle. Calculating \( Q \) gives insight into how efficiently a teakettle can bring water to a boil.
Latent Heat of Vaporization
Latent heat of vaporization is the energy required to change a substance from liquid to gas without changing its temperature. For water, this value is \( 2260 \text{ J/g} \). It explains the energy needed to boil water once it reaches boiling temperature.This concept is crucial when determining how much water evaporates from the kettle when heated. In the exercise, we used the calculated heat energy from the burner to see how much water can be converted from liquid to steam over the time given. The unexpected result of 75.7 kg of water boiled away indicates the high energy transfer rate鈥攍ikely an overestimation from idealized conditions.
Temperature Difference
Temperature difference, \( \Delta T \), is a driving force in heat conduction. It reflects how much hotter one surface is compared to another. In the exercise, the burner is at \( 150^{\circ}\text{C} \) while the kettle's water boils at \( 100^{\circ}\text{C} \), creating a \( 50^{\circ}\text{C} \) difference. This difference directly influences the rate of heat transfer. A higher \( \Delta T \) means more heat is likely to flow, speeding up boiling. Understanding \( \Delta T \) helps explain why heat transfer often slows down as temperatures even out. The teakettle exercise shows how \( \Delta T \) plays a role in potential overestimation, as constant temperatures were assumed.

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Most popular questions from this chapter

A large window measures \(2.0 \mathrm{~m}\) by \(3.0 \mathrm{~m}\). At what rate will heat be conducted through the window when the room temperature is \(20^{\circ} \mathrm{C}\) and the outside temperature is \(0^{\circ} \mathrm{C}\) if (a) the window consists of a single pane of glass \(4.0 \mathrm{~mm}\) thick and (b) the window instead has a double pane of glass (a "thermopane"), in which each pane is \(2.0 \mathrm{~mm}\) thick, with an intervening air space of \(1.0 \mathrm{~mm} ?\) (Assume that there is a constant temperature difference and consider conduction only.)

How much heat is required to boil away \(1.50 \mathrm{~kg}\) of water that is initially at \(100^{\circ} \mathrm{C} ?\)

In a calorimetry experiment, \(0.50 \mathrm{~kg}\) of a metal at \(100^{\circ} \mathrm{C}\) is added to \(0.50 \mathrm{~kg}\) of water at \(20^{\circ} \mathrm{C}\) in an aluminum calorimeter cup. The cup has a mass of \(0.250 \mathrm{~kg}\). (a) If some water splashed out of the cup when the metal was added, the measured specific heat will appear to be (1) higher, (2) the same, (3) lower than the value calculated for the case in which the water does not splash out. Why? (b) If the final temperature of the mixture is \(25^{\circ} \mathrm{C}\), and no water splashed out, what is the specific heat of the metal?

A window air conditioner has a rating of \(20000 \mathrm{Btu} / \mathrm{h}\). What is this rating in watts?

A person goes on a 1500-Cal-per-day diet to lose weight. What is his daily energy allowance expressed in joules?
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