Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 8: Problem 2

Gold has a specific heat of \(0.129 \mathrm{~J} / \mathrm{g} \cdot{ }^{\circ} \mathrm{C}\). When a \(5.00-\mathrm{g}\) piece of gold absorbs \(1.33\) J of heat, what is the change in temperature?

Short Answer

Expert verified
Answer: The change in temperature of the gold piece is 2.06°C.

Step by step solution

01

Identify the given values

The given values are: - Specific heat of gold (\(c\)) = \(0.129 \mathrm{~J} / \mathrm{g} \cdot{ }^{\circ} \mathrm{C}\) - Mass of the gold piece (\(m\)) = \(5.00\,\text{g}\) - Heat absorbed (\(q\)) = \(1.33\,\text{J}\)
02

Use the formula for heat transfer

We will use the heat transfer formula \(q=mc\Delta T\), where \(q\) is the heat transferred, \(m\) is the mass of the object, \(c\) is the specific heat, and \(\Delta T\) is the change in temperature.
03

Solve for the change in temperature \(\Delta T\)

We will rearrange the equation to solve for the change in temperature: \(\Delta T = \frac{q}{mc}\) Substitute the given values into the equation: \(\Delta T = \frac{1.33\,\text{J}}{5.00\,\text{g} \cdot 0.129 \,\mathrm{J} / \mathrm{g} \cdot{ }^{\circ} \mathrm{C}}\)
04

Calculate the change in temperature

Now, we will perform the calculation to find the change in temperature: \(\Delta T = \frac{1.33\,\text{J}}{5.00\,\text{g} \cdot 0.129 \,\mathrm{J} / \mathrm{g} \cdot{ }^{\circ} \mathrm{C}} = \frac{1.33\,\text{J}}{0.645\,\mathrm{J} /{ }^{\circ} \mathrm{C}} = 2.06 { }^{\circ} \mathrm{C}\)
05

State the final answer

The change in temperature of the gold piece when it absorbs \(1.33\,\text{J}\) of heat is \(2.06 { }^{\circ} \mathrm{C}\).

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.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

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

Nitroglycerine, \(\mathrm{C}_{3} \mathrm{H}_{5}\left(\mathrm{NO}_{3}\right)_{3}(l)\), is a powerful explosive used in rock blasting when roads are created. When ignited, it produces water, nitrogen, carbon dioxide, and oxygen. Detonation of one mole of nitroglycerine liberates \(5725 \mathrm{~kJ}\) of heat. (a) Write a balanced thermochemical equation for the reaction for the detonation of four moles of nitroglycerine. (b) What is \(\Delta H_{\mathrm{f}}^{\circ}\) for \(\mathrm{C}_{3} \mathrm{H}_{5}\left(\mathrm{NO}_{3}\right)_{3}(l) ?\)

Urea, \(\left(\mathrm{NH}_{2}\right)_{2} \mathrm{CO}\), is used in the manufacture of resins and glues. When \(5.00 \mathrm{~g}\) of urea is dissolved in \(250.0 \mathrm{~mL}\) of water \((d=1.00 \mathrm{~g} / \mathrm{mL})\) at \(30.0^{\circ} \mathrm{C}\) in a coffee-cup calorimeter, \(27.6 \mathrm{~kJ}\) of heat is absorbed. (a) Is the solution process exothermic? (b) What is \(q_{\mathrm{H}_{2} \mathrm{O}}\) ? (c) What is the final temperature of the solution? (Specific heat of water is \(4.18 \mathrm{~J} / \mathrm{g} \cdot{ }^{\circ} \mathrm{C}\).) (d) What are the initial and final temperatures in \({ }^{\circ} \mathrm{F}\) ?

Copper is used in building the integrated circuits, chips, and printed circuit boards for computers. When \(228 \mathrm{~J}\) of heat are absorbed by \(125 \mathrm{~g}\) of copper at \(22.38^{\circ} \mathrm{C}\), the temperature increases to \(27.12^{\circ} \mathrm{C}\). What is the specific heat of copper?

Titanium is a metal used in jet engines. Its specific heat is \(0.523 \mathrm{~J} / \mathrm{g} \cdot{ }^{\circ} \mathrm{C}\). If \(5.88 \mathrm{~g}\) of titanium absorbs \(4.78 \mathrm{~J}\), what is the change in temperature?

Which statement(s) is/are true about bond enthalpy? (a) Energy is required to break a bond. (b) \(\Delta H\) for the formation of a bond is always a negative number. (c) Bond enthalpy is defined only for bonds broken or formed in the gaseous state. (d) Because the presence of \(\pi\) bonds does not influence the geometry of a molecule, the presence of \(\pi\) bonds does not affect the value of the bond enthalpy between two atoms either. (e) The bond enthalpy for a double bond between atoms \(A\) and \(B\) is twice that for a single bond between atoms \(\mathrm{A}\) and \(\mathrm{B}\).
See all solutions

Recommended explanations on Chemistry Textbooks

Ionic and Molecular Compounds

Read Explanation

The Earths Atmosphere

Read Explanation

Chemistry Branches

Read Explanation

Making Measurements

Read Explanation

Chemical Reactions

Read Explanation

Chemical Analysis

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.

Sign-up for free