/*! 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 4 A truck initially traveling at \... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 6: Problem 4

A truck initially traveling at \(60 \mathrm{~km}\) per hour is brought to a complete stop at a traffic light. Does this change violate the law of conservation of energy? Explain.

Short Answer

Expert verified
No, the situation does not violate the law of conservation of energy. The kinetic energy of the truck is converted into other forms of energy, such as heat due to the brakes' friction, sound, and potentially a slight amount of potential energy, which means that energy is conserved.

Step by step solution

01

Identify initial and final state

In the beginning, the truck has a certain amount of kinetic energy due to its speed of 60 km/h. That's the initial state of the system. When it comes to a stop, the truck has zero kinetic energy. That's the final state.
02

Apply the conservation law

According to the law of conservation of energy, the energy should remain unchanged. It can be transformed into other forms, but not lost or created. The kinetic energy of the truck should be converted into some other form when it comes to a stop.
03

Explain the energy transformation

When the truck stops, the kinetic energy it had doesn't just disappear - it gets transformed into other forms of energy. The biggest part of it is probably converted into heat due to the friction between the truck's brakes and its wheels. Some of it might also get converted into sound energy, and a tiny amount into potential energy if the stop made the brakes compress slightly. Thus, there's no violation of the law of conservation of 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!

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

A 44.0-g sample of an unknown metal at \(99.0^{\circ} \mathrm{C}\) was placed in a constant-pressure calorimeter containing \(80.0 \mathrm{~g}\) of water at \(24.0^{\circ} \mathrm{C}\). The final temperature of the system was found to be \(28.4^{\circ} \mathrm{C}\). Calculate the specific heat of the metal. (The heat capacity of the calorimeter is \(\left.12.4 \mathrm{~J} /{ }^{\circ} \mathrm{C} .\right)\)

From the following data, $$ \begin{array}{c} \mathrm{C} \text { (graphite) }+\mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}(g) \\ \Delta H_{\mathrm{rxn}}^{\circ}=-393.5 \mathrm{~kJ} / \mathrm{mol} \\\ \mathrm{H}_{2}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{H}_{2} \mathrm{O}(l) \\ \Delta H_{\mathrm{rxn}}^{\circ}=-285.8 \mathrm{~kJ} / \mathrm{mol} \\ 2 \mathrm{C}_{2} \mathrm{H}_{6}(g)+7 \mathrm{O}_{2}(g) \longrightarrow 4 \mathrm{CO}_{2}(g)+6 \mathrm{H}_{2} \mathrm{O}(l) \\ \Delta H_{\mathrm{rxn}}^{\circ}=-3119.6 \mathrm{~kJ} / \mathrm{mol} \end{array} $$ calculate the enthalpy change for the reaction $$ 2 \mathrm{C}(\text { graphite })+3 \mathrm{H}_{2}(g) \longrightarrow \mathrm{C}_{2} \mathrm{H}_{6}(g) $$

The standard enthalpy change for the following reaction is \(436.4 \mathrm{~kJ} / \mathrm{mol}\) : $$ \mathrm{H}_{2}(g) \longrightarrow \mathrm{H}(g)+\mathrm{H}(g) $$ Calculate the standard enthalpy of formation of atomic hydrogen (H).

The combustion of \(0.4196 \mathrm{~g}\) of a hydrocarbon releases \(17.55 \mathrm{~kJ}\) of heat. The masses of the products are \(\mathrm{CO}_{2}=1.419 \mathrm{~g}\) and \(\mathrm{H}_{2} \mathrm{O}=0.290 \mathrm{~g} .\) (a) What is the empirical formula of the compound? (b) If the approximate molar mass of the compound is \(76 \mathrm{~g}\), calculate its standard enthalpy of formation.

Stoichiometry is based on the law of conservation of mass. On what law is thermochemistry based?
See all solutions

Recommended explanations on Chemistry Textbooks

Physical Chemistry

Read Explanation

Chemical Analysis

Read Explanation

Chemical Reactions

Read Explanation

Making Measurements

Read Explanation

The Earths Atmosphere

Read Explanation

Ionic and Molecular Compounds

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.