/*! 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 64 It's possible to boil water by a... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 12: Problem 64

It's possible to boil water by adding hot rocks to it, a technique that has been used in many societies over time. If you heat a rock in the fire, you can easily get it to a temperature of \(500^{\circ} \mathrm{C} .\) If you use granite or other similar stones, the specific heat is about \(800 \mathrm{J} / \mathrm{kg} \cdot \mathrm{K}\). If \(5.0 \mathrm{kg}\) of water at \(10^{\circ} \mathrm{C}\) is in a leak-proof vessel, what minimum number of \(1.0 \mathrm{kg}\) stones must be added to bring the water to a boil?

Short Answer

Expert verified
In order to get the final numerical answer, fill in the numerical values calculated in previous steps into the formula for the number of stones and perform the division.

Step by step solution

01

Calculate the heat required to boil the water

First, we will calculate the heat needed to raise the temperature of water from 10 degrees Celsius to 100 degrees Celsius (the boiling point of water). The formula \(Q = mc\Delta T\) is used, where Q is the heat transferred, m is the mass (5.0 kg for water), c is the specific heat (approximately 4200 J/kg.K for water), and \(\Delta T\) is the change in temperature (100 - 10 = 90 degrees Celsius). Thus, \(Q_{water} = 5.0 kg * 4200 J/kg.K * 90 K\).
02

Calculate the heat transferred by each stone

Next, we calculate the heat lost by a single 1 kg stone from 500 degrees Celsius to 100 degrees Celsius. Using the same formula, \(Q = mc\Delta T\), where m is the mass (1 kg for the stone), c is the specific heat (approximately 800 J/kg.K for stone), and \(\Delta T\) is the change in temperature (500 - 100 = 400 degrees Celsius), we find \(Q_{stone} = 1 kg * 800 J/kg.K * 400 K\).
03

Find number of stones

Finally, the number of stones needed to provide the required heat to the water is obtained by dividing the total heat required by the water by the heat provided by each stone, that is, Number of stones = \(Q_{water} / Q_{stone}\).

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.

Heat Transfer
Heat transfer is the process by which thermal energy moves from a hotter object to a cooler one. Heat always flows in this direction naturally, which is why in the exercise, hot rocks can be used to boil water. There are three primary methods of heat transfer: conduction, convection, and radiation.

Conduction occurs when heat is transferred through direct contact, as between the hot rocks and water in the exercise.
Convection involves the movement of heat through fluids, like water, as warmer areas rise and cooler areas sink.
Radiation is the transfer of heat through electromagnetic waves, which is less relevant to this scenario.

In the exercise, the heat from the rocks is conducted into the water, raising its temperature. This process continues until thermal equilibrium is achieved, meaning the temperature of the rocks and water equalizes.
Specific Heat Capacity
Specific heat capacity is a property of matter that tells us how much heat is needed to raise the temperature of a unit mass of a substance by one degree Celsius (or Kelvin). Different materials have different specific heat capacities.

The formula to calculate the heat transferred is: \[ Q = mc\Delta T \]where:
  • \(Q\) is the heat transferred,
  • \(m\) is the mass,
  • \(c\) is the specific heat capacity,
  • \(\Delta T\) is the change in temperature.


In the exercise, the specific heat capacity of water is given as approximately 4200 J/kg鈰匥 and for granite stones, it is 800 J/kg鈰匥. This means water requires more energy to increase in temperature than granite stones do for the same mass. Knowing the specific heat capacity is crucial in calculating how much heat energy is involved in heating substances.
Temperature Change
Temperature change can influence many physical properties and processes, and it is a core aspect of thermodynamics. In our case, calculating how much the temperature changes is vital to deciding how much energy is needed or will be exchanged.

When dealing with heat transfer as in our exercise, the temperature change (\(\Delta T\)) is the difference between the final and initial temperature. For water, it's from 10掳C to 100掳C; and for rocks, from 500掳C down to 100掳C.

This change allows us to compute the amount of energy required or released using the formula \( Q = mc\Delta T \). Temperature change is a direct indicator of the thermal energy involved in boiling the water.
Energy Conservation
Energy conservation, a principal law of physics, states that energy cannot be created or destroyed--only transformed from one form to another. When considering heat transfer, this means the total energy in a closed system must remain constant.

In the boiling water exercise, the energy lost by the hot rocks must equal the energy gained by the water if there is no loss to the environment.
  • The heat energy lost by the rocks as they cool down is transferred to the water.
  • This raises the water鈥檚 temperature until it reaches its boiling point.
  • If any energy is lost to the surroundings, more rocks would be needed.


Understanding energy conservation allows us to calculate the number of rocks required to boil water, assuming perfect transfer efficiency.

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

Many fish maintain buoyancy with a gas-filled swim bladder. The pressure inside the swim bladder is the same as the outside water pressure, so when a fish descends to a greater depth, the gas compresses. Adding gas to restore the original volume requires energy. A fish at a depth where the absolute pressure is 3.0 atm has a swim bladder with the desired volume of \(5.0 \times 10^{-4} \mathrm{m}^{3} .\) The fish now descends to a depth where the absolute pressure is 5.0 atm. a. The gas in the swim bladder is always the same temperature as the fish's body. What is the volume of the swim bladder at the greater depth? b. The fish remains at the greater depth, slowly adding gas to the swim bladder to return it to its desired volume. How much work is required?

The pressure inside a champagne bottle can be quite high and can launch a cork explosively. Suppose you open a bottle at sea level. The absolute pressure inside a champagne bottle is 6 times atmospheric pressure; the cork has a mass of 7.5 g and a diameter of \(18 \mathrm{mm}\). Assume that once the cork starts to move, the only force that matters is the pressure force. What is the acceleration of the cork?

\(0.10 \mathrm{mol}\) of argon gas is admitted to an evacuated \(50 \mathrm{cm}^{3}\) container at \(20^{\circ} \mathrm{C}\). The gas then undergoes an isobaric heating to a temperature of \(300^{\circ} \mathrm{C}\). a. What is the final volume of the gas? b. Show the process on a \(p V\) diagram. Include a proper scale on both axes.

A 30 kg male emperor penguin under a clear sky in the Antarctic winter loses very little heat to the environment by convection; its feathers provide very good insulation. It does lose some heat through its feet to the ice, and some heat due to evaporation as it breathes; the combined power is about \(12 \mathrm{W}\). The outside of the penguin's body is a chilly \(-22^{\circ} \mathrm{C},\) but its surroundings are an even chillier \(-38^{\circ} \mathrm{C} .\) The penguin's surface area is \(0.56 \mathrm{m}^{2},\) and its emissivity is 0.97. a. What is the net rate of energy loss by radiation? b. If the penguin has a 45 W basal metabolic rate, will it feel warm or cold under these circumstances?

An expandable cube, initially \(20 \mathrm{cm}\) on each side, contains \(3.0 \mathrm{g}\) of helium at \(20^{\circ} \mathrm{C} .1000 \mathrm{J}\) of heat energy are transferred to this gas. What are (a) the final pressure if the process is at constant volume and (b) the final volume if the process is at constant pressure?
See all solutions

Recommended explanations on Physics Textbooks

Electricity and Magnetism

Read Explanation

Turning Points in Physics

Read Explanation

Dynamics

Read Explanation

Wave Optics

Read Explanation

Electrostatics

Read Explanation

Translational Dynamics

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.