/*! 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 53 A pot with a steel bottom 8.50 \... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 14: Problem 53

A pot with a steel bottom 8.50 \(\mathrm{mm}\) thick rests on a hot stove. The area of the bottom of the pot is 0.150 \(\mathrm{m}^{2} .\) The water inside the pot is at \(100.0^{\circ} \mathrm{C},\) and 0.390 \(\mathrm{kg}\) are evaporated every 3.00 min. Find the temperature of the lower surface of the pot, which is in contact with the stove.

Short Answer

Expert verified
The temperature of the lower surface of the pot is approximately 123.21掳C.

Step by step solution

01

Identify Given Information

We have a steel pot with a bottom thickness of 8.50 mm, which we convert to meters as 0.00850 m. The bottom area of the pot is 0.150 m虏. Water evaporates at a rate of 0.390 kg every 3.00 minutes at a boiling point of 100.0掳C.
02

Calculate Heat Needed to Evaporate Water

The heat required to vaporize the water is given by the equation \( Q = mL \), where \( m = 0.390 \text{ kg} \) is the mass, and \( L = 2.26 \times 10^6 \text{ J/kg} \) is the latent heat of vaporization for water. Thus, \( Q = 0.390 \times 2.26 \times 10^6 = 881400 \text{ J} \) in 3.00 minutes.
03

Find Heat Transfer Rate

To find the heat transfer rate \( \dot{Q} \), we divide the total heat by time: \( \dot{Q} = \frac{Q}{t} \), where \( t = 3 \times 60 = 180 \text{ s} \). Therefore, \( \dot{Q} = \frac{881400}{180} \approx 4897.22 \text{ W} \).
04

Apply Fourier's Law of Heat Conduction

Fourier's Law is \( \dot{Q} = \frac{kA(T_{surface} - T_{water})}{d} \), where \( k \) for steel is approximately 50.2 W/m路K. Plugging in the values: \( 4897.22 = \frac{(50.2)(0.150)(T_{surface} - 100)}{0.0085} \).
05

Solve for Temperature of Lower Surface

Rearrange to solve for \( T_{surface} \):\[ T_{surface} = \frac{4897.22 \times 0.0085}{50.2 \times 0.150} + 100. \] Calculating the expression gives \( T_{surface} \approx 123.21^{\circ} \text{C} \).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Fourier's Law
Fourier's Law of heat conduction is a fundamental principle in thermal physics. It describes how heat energy is transferred through a material due to a temperature difference. According to Fourier's Law, the rate of heat transfer (\( \dot{Q} \)) through a material is proportional to the negative gradient of temperature and the area perpendicular to the gradient. The formula can be expressed as:
\[\dot{Q} = -kA \frac{dT}{dx}\]Here, \( k \) is the thermal conductivity of the material, \( A \) is the area through which heat is conducted, and \( \frac{dT}{dx} \) is the temperature gradient.
  • In the exercise, Fourier's Law helps us calculate how much heat is transferred through the steel bottom of the pot.
  • It involves a simple rearrangement to solve for the temperature at one side of the steel sheet.
  • The equation allows us to understand the influence of material properties like thermal conductivity on heat transfer.
latent heat of vaporization
The latent heat of vaporization is the heat required to convert a unit mass of a liquid into vapor without a change in temperature. This concept is crucial when studying processes like boiling and evaporation. In terms of physics, it involves the breaking of molecular bonds in a liquid state to form a gas.
  • The latent heat of vaporization for water is approximately \( 2.26 \times 10^6 \) J/kg.
  • This value indicates the amount of energy needed to turn 1 kilogram of water into steam, maintaining a constant temperature of 100掳C.
  • In the exercise, this energy conversion is calculated to understand how much heat is absorbed from the stove to vaporize the water in the pot, which is measured over a period of 3 minutes.
Understanding the latent heat of vaporization helps in calculating energy changes in heating processes and designing heating systems and equipment.
thermal conductivity
Thermal conductivity (\( k \)) is a property of a material that determines how well it can conduct heat. Higher thermal conductivity means the material is a better conductor of heat. It's a crucial factor in calculations involving heat transfer through solids.
  • For example, steel, which is used in the pot in this exercise, has a thermal conductivity of approximately \( 50.2 \) W/m路K.
  • Materials with high thermal conductivity transfer heat more efficiently compared to insulating materials with lower values of \( k \).
  • In the given problem, we use this property of steel to find how heat flows from the stove through the pot to the boiling water.
Having knowledge about thermal conductivity allows engineers and scientists to design systems that manage temperature and energy consumption effectively.
temperature difference
The concept of temperature difference is essential in understanding heat flow. Heat naturally flows from areas of higher temperature to areas of lower temperature. The magnitude of this temperature difference greatly affects the rate of heat transfer.
  • In the exercise, the temperature difference between the stove (lower surface) and the boiling water (upper surface) drives the conduction process.
  • Calculating the temperature difference is crucial in determining the heat transfer rate as per Fourier's Law.
  • Analyzing temperature differences helps us set the necessary conditions for achieving desired thermal balance in various industrial applications.
Temperature difference is a simple yet powerful concept that allows one to predict and manipulate heat flow effectively.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

\(\bullet\) The markings on an aluminum ruler and a brass ruler begin at the left end; when the rulers are at \(0.00^{\circ} \mathrm{C}\) , they are perfectly aligned. How far apart will the 20.0 \(\mathrm{cm}\) marks be on the two rulers at \(100.0^{\circ} \mathrm{C}\) if the left-hand ends are kept precisely aligned?

". "The Ship of the Desert." Camels require very little water because they are able to tolerate relatively large changes in their body temperature. While humans keep their body temperatures constant to within one or two Celsius degrees, a dehydrated camel permits its body temperature to drop to \(34.0^{\circ} \mathrm{C}\) overnight and rise to \(40.0^{\circ} \mathrm{C}\) during the day. To see how effective this mechanism is for saving water, calculate how many liters of water a 400 -kg camel would have to drink if it attempted to keep its body temperature at a constant \(34.0^{\circ} \mathrm{C}\) by evaporation of sweat during the day \((12\) hours) instead of letting it rise to \(40.0^{\circ} \mathrm{C}\) . (Note: The specific heat of a camel or other mammal is about the same as that of a typical human, 3480 \(\mathrm{J} /(\mathrm{kg} \cdot \mathrm{K}) .\) The heat of vaporization of water at \(34^{\circ} \mathrm{C}\) is \(2.42 \times 10^{6} \mathrm{J} / \mathrm{kg} . )\)

A technician measures the specific heat capacity of an unidentified liquid by immersing an electrical resistor in it. Electrical energy is converted to heat, which is then transferred to the liquid for 120 s at a constant rate of 65.0 W. The mass of the liquid is \(0.780 \mathrm{kg},\) and its temperature increases from \(18.55^{\circ} \mathrm{C}\) to \(22.54^{\circ} \mathrm{C}\) . (a) Find the average specific heat capacity of the liquid in this temperature range. Assume that negligible heat is transferred to the container that holds the liquid and that no heat is lost to the surroundings. (b) Suppose that in this experiment heat transfer from the liquid to the container or its surroundings cannot be ignored. Is the result calculated in part (a) an overestimate or an underestimate of the average specific heat capacity? Explain.

Shivering. You have no doubt noticed that you usually shiver when you get out of the shower. Shivering is the body's way of generating heat to restore its internal temperature to the normal \(37^{\circ} \mathrm{C},\) and it produces approximately 290 \(\mathrm{W}\) of heat power per square meter of body area. \(\mathrm{A} 68 \mathrm{kg}(150 \mathrm{lb}), 1.78 \mathrm{m}\) (5 foot, 10 inch) person has approximately 1.8 \(\mathrm{m}^{2}\) of surface area. How long would this person have to shiver to raise his or her body temperature by \(1.0 \mathrm{C}^{\circ},\) assuming that none of this heat is lost by the body? The specific heat capacity of the body is about 3500 \(\mathrm{J} /(\mathrm{kg} \cdot \mathrm{K})\) .

\(\bullet\) A 0.500 kg chunk of an unknown metal that has been in boiling water for several minutes is quickly dropped into an insulating Styrofoam TM beaker containing 1.00 kg of water at room temperature \(\left(20.0^{\circ} \mathrm{C}\right) .\) After waiting and gently stirring for 5.00 minutes, you observe that the water's temperature has reached a constant value of \(22.0^{\circ} \mathrm{C}\) (a) Assuming that the Styrofoam absorbs a negligibly small amount of heat and that no heat was lost to the surroundings, what is the specific heat capacity of the metal? (b) Which is more useful for storing energy from heat, this metal or an equal weight of water? Explain. (c) What if the heat absorbed by the Styrofoam"" actually is not negligible. How would the specific heat capacity you calculated in part (a) be in error? Would it be too large, too small, or still correct? Explain your reasoning.
See all solutions

Recommended explanations on Physics Textbooks

Energy Physics

Read Explanation

Electricity

Read Explanation

Force

Read Explanation

Linear Momentum

Read Explanation

Thermodynamics

Read Explanation

Fluids

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.