/*! 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 15 A pot of water is boiling under ... [FREE SOLUTION] | 91Ó°ÊÓ

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A pot of water is boiling under one atmosphere of pressure. Assume that heat enters the pot only through its bottom, which is copper and rests on a heating element. In two minutes, the mass of water boiled away is \(m=0.45 \mathrm{kg} .\) The radius of the pot bottom is \(R=6.5 \mathrm{cm},\) and the thickness is \(L=2.0 \mathrm{mm} .\) What is the temperature \(T_{\mathrm{E}}\) of the heating element in contact with the pot?

Short Answer

Expert verified
The temperature of the heating element is approximately 129°C.

Step by step solution

01

Understand the Problem

We need to find the temperature \( T_{\mathrm{E}} \) of the heating element in contact with the pot using the given values: \( m = 0.45 \, \text{kg}\), \( R = 6.5 \, \text{cm} = 0.065 \, \text{m}\), and \( L = 2.0 \, \text{mm} = 0.002 \, \text{m}\). The latent heat of vaporization for water is \( L_v = 2260 \, \text{kJ/kg} \).
02

Calculate Heat Required to Boil Water

To calculate the heat \( Q \) required to boil 0.45 kg of water, use the formula \( Q = m \, L_v \). Substitute the values to get \( Q = 0.45 \, \text{kg} \times 2260 \, \text{kJ/kg} \), resulting in \( Q = 1017 \, \text{kJ} = 1017 \times 10^3 \, \text{J} \).
03

Determine Heat Transfer Rate through Copper

Find the rate of heat transfer \( P \) from the heating element to the water using the formula \( P = \frac{Q}{\Delta t} \). Given \( \Delta t = 2 \, \text{minutes} = 120 \, \text{seconds} \), we have \( P = \frac{1017 \times 10^3 \, \text{J}}{120 \, \text{s}} \approx 8475 \, \text{W} \).
04

Use Fourier’s Law of Thermal Conduction

Use Fourier’s Law to find the temperature difference \( \Delta T \) across the copper: \( P = \frac{k \pi R^2 (T_{\mathrm{E}} - 100)}{L} \), where \( k \) is the thermal conductivity of copper (about 385 \( \text{W/m} \, \text{°C} \)).
05

Calculate the Temperature of the Heating Element

Rearrange Fourier’s Law to solve for \( T_{\mathrm{E}} \): \( T_{\mathrm{E}} = \frac{P \times L}{k \times \pi R^2} + 100 \). Substitute the known values: \( T_{\mathrm{E}} = \frac{8475 \, \text{W} \times 0.002 \, \text{m}}{385 \, \text{W/m°C} \times \pi \times (0.065 \, \text{m})^2} + 100 \).
06

Final Calculation

Perform the calculation: \( T_{\mathrm{E}} \approx \frac{8475 \times 0.002}{385 \times \pi \times 0.065^2} + 100 \approx 128.85 \, \text{°C} \). Therefore, \( T_{\mathrm{E}} \approx 129 \, \text{°C} \).

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Key Concepts

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

Fourier's Law of Thermal Conduction
Fourier's Law of Thermal Conduction is a cornerstone concept in understanding how heat moves through materials. It explains how the rate of heat transfer (denoted as \( P \)) through a solid surface is related to the thermal properties of the material itself. This is particularly important for substances like copper, which exhibit high thermal conductivity.

In mathematical terms, Fourier's Law is expressed as:
  • \( P = \frac{k A (T_1 - T_2)}{L} \)
  • where \( P \) is the heat transfer rate, \( k \) is the thermal conductivity, \( A \) is the cross-sectional area, \( T_1 \) and \( T_2 \) are the temperatures on either side of the material, and \( L \) is the thickness of the material.
Understanding this law helps us calculate how efficiently heat can be transferred through the copper bottom of the pot.

In our exercise, we apply Fourier's law to find the unknown temperature on one side of the copper—the temperature of the heating element—even if we know the heat rate, conductivity, and the size of the conducting cross-section.
Boiling Point
The boiling point of a substance is the temperature at which it changes from a liquid to a gas. For water, this occurs at 100°C under one atmosphere of pressure.

Why is boiling point important? It serves as the baseline temperature in our exercise. Once water reaches its boiling point, any additional heat supplied doesn't increase the water's temperature. Instead, it contributes to the phase change from liquid to vapor.

In practical scenarios, like boiling water in a pot, understanding the boiling point helps us predict and control the energy input required for vaporization. Since the boiling point remains constant while a liquid is being converted to gas, it plays a crucial role in calculating other heat-related parameters in the system.
Heat Transfer
Heat transfer involves the movement of thermal energy from one body to another. There are three modes of heat transfer: conduction, convection, and radiation. In this exercise, we focus on conduction, which is the transfer of heat through a material without the material itself moving.

The purpose of calculating heat transfer in the problem is to estimate how efficiently the heat from the element flows through the copper bottom and into the water. This is crucial for finding the temperature of the heating element. The rate of heat transfer in watts is essential for ensuring that the right amount of energy is provided over a given time to achieve the desired phase change for water.

With an accurate understanding of heat transfer, especially conduction, we can effectively design and simulate real-world systems where temperature control is vital, like cooking or industrial processes.
Latent Heat of Vaporization
Latent heat of vaporization is the amount of energy required to transform a given quantity of substance from a liquid into a gas at constant temperature and pressure. For water, the latent heat of vaporization is significant, at 2260 kJ/kg.

In the exercise, this concept is pivotal as it dictates the energy necessary to boil 0.45 kg of water. Given the mass of the water and the latent heat, we can determine the total energy input required to evaporate this mass completely. Without an understanding of latent heat, calculating the energy requirements for phase changes would be challenging.

This energy does not increase the temperature of the water; it only contributes to the phase transition. Thus, latent heat is a key parameter in thermodynamics and thermal calculations related to phase changes, ensuring accuracy and efficiency in processes involving boiling and condensation.

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Most popular questions from this chapter

One half of a kilogram of liquid water at \(273 \mathrm{K}\left(0^{\circ} \mathrm{C}\right)\) is placed outside on a day when the temperature is \(261 \mathrm{K}\left(-12^{\circ} \mathrm{C}\right) .\) Assume that the heat is lost from the water only by means of radiation and that the emissivity of the radiating surface is \(0.60 .\) Consider two cases: when the surface area of the water is (a) \(0.035 \mathrm{m}^{2}\) (as it might be in a cup) and (b) \(1.5 \mathrm{m}^{2}\) (as it could be if the water were spilled out to form a thin sheet). Concepts: (i) In case (a) is the heat that must be removed to freeze the water less than, greater than, or the same as in case (b)? (ii) The loss of heat by radiation depends on the temperature of the radiating object. Does the temperature of the water change as it freezes? (iii) The water both loses and gains heat by radiation. How, then, can heat transfer by radiation lead to freezing of the water? (iv) Will it take longer for the water to freeze in case (a) when the area is smaller or in case (b) when the area is larger? Calculations: For each case, (a) and (b), determine the time it takes the water to freeze into ice at \(0^{\circ} \mathrm{C}\).

Sirius \(\mathrm{B}\) is a white star that has a surface temperature (in kelvins) that is four times that of our sun. Sirius \(\mathrm{B}\) radiates only 0.040 times the power radiated by the sun. Our sun has a radius of \(6.96 \times 10^{8} \mathrm{m} .\) Assuming that Sirius B has the same emissivity as the sun, find the radius of Sirius B.

One end of an iron poker is placed in a fire where the temperature is \(502^{\circ} \mathrm{C}\), and the other end is kept at a temperature of \(26^{\circ} \mathrm{C}\). The poker is \(1.2 \mathrm{m}\) long and has a radius of \(5.0 \times 10^{-3} \mathrm{m} .\) Ignoring the heat lost along the length of the poker, find the amount of heat conducted from one end of the poker to the other in \(5.0 \mathrm{s}\).

A closed box is filled with dry ice at a temperature of \(-78.5^{\circ} \mathrm{C}\), while the outside temperature is \(21.0^{\circ} \mathrm{C} .\) The box is cubical, measuring \(0.350 \mathrm{m}\) on a side, and the thickness of the walls is \(3.00 \times 10^{-2} \mathrm{m} .\) In one day, \(3.10 \times 10^{6} \mathrm{J}\) of heat is conducted through the six walls. Find the thermal conductivity of the material from which the box is made.

In an old house, the heating system uses radiators, which are hollow metal devices through which hot water or steam circulates. In one room the radiator has a dark color (emissivity \(=0.75\) ). It has a temperature of \(62^{\circ} \mathrm{C} .\) The new owner of the house paints the radiator a lighter color (emissivity \(=0.50\) ). Assuming that it emits the same radiant power as it did before being painted, what is the temperature (in degrees Celsius) of the newly painted radiator?

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