/*! 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 73 A bag containing \(0^{\circ} \ma... [FREE SOLUTION] | 91影视

91影视

A bag containing \(0^{\circ} \mathrm{C}\) ice is much more effective in absorbing energy than one containing the same amount of \(0^{\circ} \mathrm{C}\) water. (a) How much heat transfer is necessary to raise the temperature of \(0.800 \mathrm{kg}\) of water from \(0^{\circ} \mathrm{C}\) to \(30.0^{\circ} \mathrm{C} ?\) (b) How much heat transfer is required to first melt \(0.800 \mathrm{kg}\) of \(0^{\circ} \mathrm{C}\) ice and then raise its temperature? (c) Explain how your answer supports the contention that the ice is more effective.

Short Answer

Expert verified
To raise the temperature of 0.800 kg of water from 0鈩 to 30鈩, 100,480 J of heat transfer is required. For 0.800 kg of 0鈩 ice, first melting the ice requires 266,400 J of heat transfer, and then raising its temperature requires another 100,480 J, totaling 366,880 J. The ice absorbs about 3.65 times more energy than water, making it more effective in absorbing energy.

Step by step solution

01

Identify known values for water

We know the following for the given water: Mass (m) = 0.800 kg Initial temperature (T鈧) = 0鈩 Final temperature (T鈧) = 30.0鈩 Specific heat capacity of water (c) = 4186 J/kg鈩
02

Calculate heat transfer for water

Use the formula Q = mc螖罢 to calculate the heat transfer: 螖罢 = T鈧 - T鈧 = 30鈩 - 0鈩 = 30鈩 Q = (0.800 kg) (4186 J/kg鈩) (30鈩) Q = 100480 J So, 100,480 J of heat transfer is required to raise the temperature of the water. #b) Calculate heat transfer to melt ice and raise water temperature#
03

Identify known values for ice

We know the following for the given ice: Mass (m) = 0.800 kg Latent heat of fusion of ice (L) = 333,000 J/kg Specific heat capacity of water (c) = 4186 J/kg鈩 (since we are raising the temperature of water after melting the ice)
04

Calculate heat transfer to melt the ice

Use the formula Q = mL to calculate the heat transfer to melt the ice: Q = (0.800 kg)(333,000 J/kg) Q = 266400 J
05

Calculate heat transfer to raise water temperature

Use the same method as in part (a) to find the heat transfer to raise the temperature: Q = (0.800 kg) (4186 J/kg鈩) (30鈩) Q = 100480 J
06

Calculate total heat transfer

Add heat transfer from melting ice and raising the temperature of water: Total heat transfer = Heat transfer for melting + Heat transfer for temperature rise Total heat transfer = 266400 J + 100480 J Total heat transfer = 366880 J #c) Compare heat transfers and support the contention# The heat transfer required to first melt the 0鈩 ice and then raise its temperature is 366,880 J, whereas 100,480 J is required to raise the temperature of the same amount of 0鈩 water. The ice absorbs more energy (almost 3.65 times) than the water during the entire process, making it more effective in absorbing energy.

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.

Latent Heat of Fusion
When it comes to understanding how substances change phases, the concept of latent heat of fusion plays a critical role. Latent heat of fusion is the amount of energy needed to change a substance from solid to liquid at its melting point without changing its temperature. This energy breaks the bonds between the solid molecules, allowing them to move freely as a liquid.

For instance, ice at 0掳C will not immediately turn into water at 0掳C once it starts receiving heat. Instead, it will absorb a significant amount of energy to overcome the latent heat of fusion before it can transform into liquid water, with that energy going into breaking the bonds rather than raising the temperature. For ice, this energy is quantified as 333,000 Joules per kilogram (J/kg).
Specific Heat Capacity
Understanding how much energy is required to change the temperature of a substance involves another crucial concept, known as specific heat capacity. This is a measure of the amount of heat needed to raise the temperature of one kilogram of a substance by one degree Celsius (1掳C). It's an intrinsic property of each material that tells us how resistant it is to temperature change.

Water, for example, has a high specific heat capacity of 4186 J/kg掳C. This property signifies that water can absorb a lot of heat before its temperature increases significantly. Consequently, when heating water, you need a considerable amount of energy to observe a noticeable temperature rise, making it an exceptional substance for thermal applications like heating systems or cooling bags.
Temperature Change Thermodynamics
The principles of temperature change in thermodynamics describe how heat transfer affects the temperature of an object. According to these principles, when an object absorbs heat, its temperature rises, and when it releases heat, its temperature falls. But this is only true when the object is not undergoing a phase change.

The relationship between heat transfer and temperature change can be described mathematically by the equation Q = mc螖罢, where Q represents the heat transfer, m is the mass of the substance, c is the specific heat capacity, and 螖罢 is the change in temperature. This equation is used to calculate the amount of energy required to raise the temperature of a substance after accounting for its specific heat capacity.
Energy Absorption in Phase Changes
Phase changes involve a substance changing from one state of matter to another鈥攍ike ice melting into water or water evaporating into steam. These processes require or release energy without changing the temperature of the substance. This is because the energy absorbed or released during a phase change is used to alter the bonds between the molecules rather than to increase the kinetic energy, which would raise the temperature.

For example, when 0掳C ice melts to 0掳C water, the process consumes a significant amount of energy, known as the latent heat of fusion, to change its phase. Similarly, when water boils, the energy it absorbs is known as the latent heat of vaporization. Understanding energy absorption during phase changes is essential for explaining phenomena like why ice at 0掳C is a more effective coolant than the same amount of water at the same temperature; it has the capacity to absorb additional energy during the phase change.

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

How does the latent heat of fusion of water help slow the decrease of air temperatures, perhaps preventing temperatures from falling significantly below \(0^{\circ} \mathrm{C},\) in the vicinity of large bodies of water?

When heat transfers into a system, is the energy stored as heat? Explain briefly.

Following vigorous exercise, the body temperature of an \(80.0 \mathrm{kg}\) person is \(40.0^{\circ} \mathrm{C} .\) At what rate in watts must the person transfer thermal energy to reduce the body temperature to \(37.0^{\circ} \mathrm{C}\) in \(30.0 \mathrm{min}\), assuming the body continues to produce energy at the rate of \(150 \mathrm{W}\) ? (1 watt \(=1\) joule/second or \(1 \mathrm{W}=1 \mathrm{J} / \mathrm{s}\) )

In some countries, liquid nitrogen is used on dairy trucks instead of mechanical refrigerators. A 3.00-hour delivery trip requires 200 L of liquid nitrogen, which has a density of \(808 \mathrm{kg} / \mathrm{m}^{3}\). (a) Calculate the heat transfer necessary to evaporate this amount of liquid nitrogen and raise its temperature to \(3.00^{\circ} \mathrm{C}\). (Use \(c_{\mathrm{P}}\) and assume it is constant over the temperature range.) This value is the amount of cooling the liquid nitrogen supplies. (b) What is this heat transfer rate in kilowatt- hours? (c) Compare the amount of cooling obtained from melting an identical mass of \(0-^{\circ} \mathrm{C}\) ice with that from evaporating the liquid nitrogen.

Calculate the length of a 1 -meter rod of a material with thermal expansion coefficient \(\alpha\) when the temperature is raised from 300 K to 600 K. Taking your answer as the new initial length, find the length after the rod is cooled back down to \(300 \mathrm{K}\). Is your answer 1 meter? Should it be? How can you account for the result you got?

See all solutions

Recommended explanations on Physics Textbooks

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.