Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 10: Problem 124

A 20.0 -g sample of ice at \(-10.0^{\circ} \mathrm{C}\) is mixed with 100.0 g water at \(80.0^{\circ} \mathrm{C}\) . Calculate the final temperature of the mixture assuming no heat loss to the surroundings. The heat capacities of \(\mathrm{H}_{2} \mathrm{O}(s)\) and \(\mathrm{H}_{2} \mathrm{O}(l)\) are 2.03 and \(4.18 \mathrm{J} / \mathrm{g} \cdot^{\circ} \mathrm{C},\) respectively, and the enthalpy of fusion for ice is 6.02 \(\mathrm{kJ} / \mathrm{mol} .\)

Short Answer

Expert verified
The final temperature of the mixture is \(63.0^{\circ} \mathrm{C}\).

Step by step solution

01

Find the mass of the ice and water in grams

The mass of ice is given as 20.0 g, and the mass of water is given as 100.0 g.
02

Calculate the amount of moles of ice

As 1 mole of H2O is approximately 18 grams, we will convert the mass of ice to moles. Moles of ice = \(\frac{20.0\,\text{g}}{18\,\frac{\text{g}}{\text{mol}}}\) = 1.111 mol
03

Calculate the heat needed to raise the temperature of ice to \(0^{\circ} \mathrm{C}\)

Using the heat capacity of ice (2.03 J/g°C), we can find the heat needed (q) to raise the temperature: q = mass of ice × heat capacity of ice × ΔT q = \(20.0\,\text{g} \times 2.03\,\frac{\text{J}}{\text{g} \cdot ^{\circ}\mathrm{C}} \times 10\,^{\circ}\mathrm{C}\) q = 406 J
04

Calculate the heat needed to melt the ice at \(0^{\circ} \mathrm{C}\)

We will use the enthalpy of fusion for ice (6.02 kJ/mol) to find the heat needed for melting the ice: Heat needed = moles of ice × enthalpy of fusion Heat needed = \(1.111\,\text{mol} \times 6.02\,\frac{\text{kJ}}{\text{mol}}\) Heat needed = 6.688 kJ = 6688 J
05

Combine the heat needed to increase the temperature of ice and melt it

Total heat needed for ice = heat needed to raise the temperature of ice + heat needed to melt the ice Total heat needed for ice = 406 J + 6688 J = 7094 J
06

Calculate the heat lost by the water

Using the heat capacity of water (4.18 J/g°C), we can find the heat lost by water: Heat lost by water = mass of water × heat capacity of water × ΔT Heat lost by water = \(100.0\,\text{g} \times 4.18\,\frac{\text{J}}{\text{g} \cdot^{\circ}\mathrm{C}} \times (80.0^{\circ} \mathrm{C} - T_f)\)
07

Equate the heat gained by ice to the heat lost by water and solve for final temperature

Total heat needed for ice = Heat lost by water 7094 J = \(100.0\,\text{g} \times 4.18\,\frac{\text{J}}{\text{g} \cdot^{\circ}\mathrm{C}} \times (80.0^{\circ} \mathrm{C} - T_f)\) Now, we can solve for the final temperature, \(T_f\): \(T_f = 80.0^{\circ} \mathrm{C} - \frac{7094\,\text{J}}{100.0\,\text{g} \times 4.18\,\frac{\text{J}}{\text{g} \cdot^{\circ}\mathrm{C}}}\) \(T_f = 80.0^{\circ} \mathrm{C} - 17.0^{\circ} \mathrm{C}\)
08

Calculate the final temperature of the mixture

The final temperature of the mixture is: \(T_f = 80.0^{\circ} \mathrm{C} - 17.0^{\circ} \mathrm{C}\) = \(63.0^{\circ} \mathrm{C}\) Hence, the final temperature of the mixture is \(63.0^{\circ} \mathrm{C}\).

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

Identify the most important types of interparticle forces present in the solids of each of the following substances. a. \(\mathrm{BaSO}_{4}\) b. \(\mathrm{H}_{2} \mathrm{S}\) c. \(\mathrm{Xe}\) d. \(\mathrm{C}_{2} \mathrm{H}_{6}\) e. \(\mathrm{CsI}\) f. \(P_{4}\) g. \(\mathrm{NH}_{3}\)

Rationalize why chalk (calcium carbonate) has a higher melting point than motor oil (composed of large compounds containing only carbon and hydrogen), which has a higher melting point than water (a compound that exhibits hydrogen bonding).

Which gas, \(\mathrm{CO}\) or \(\mathrm{N}_{2},\) is expected to behave more ideally at 10 \(\mathrm{atm}\) and \(-50^{\circ} \mathrm{C} ?\)

Pyrolusite is a mineral containing manganese ions and oxide ions. Its structure can best be described as a body-centered cubic array of manganese ions with two oxide ions inside the unit cell and two oxide ions each on two faces of the cubic unit cell. What is the charge on the manganese ions in pyrolusite?

In each of the following groups of substances, pick the one that has the given property. Justify your answer. a. highest boiling point: HBr, Kr, or \(\mathrm{Cl}_{2}\) b. highest freezing point: \(\mathrm{H}_{2} \mathrm{O}, \mathrm{NaCl}\) , or HF c. lowest vapor pressure at $25^{\circ} \mathrm{C} : \mathrm{Cl}_{2}, \mathrm{Br}_{2},\( or \)\mathrm{I}_{2}$ d. lowest freezing point: \(\mathrm{N}_{2}, \mathrm{CO},\) or \(\mathrm{CO}_{2}\) e. lowest boiling point: \(\mathrm{CH}_{4}, \mathrm{CH}_{3} \mathrm{CH}_{3},\) or \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{3}\) f. highest boiling point: \(\mathrm{HF}, \mathrm{HCl},\) or \(\mathrm{HBr}\) g. lowest vapor pressure at $25^{\circ} \mathrm{C} : \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{3}, \mathrm{CH}_{3} \mathrm{CCH}_{3}$ or \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{OH}\)
See all solutions

Recommended explanations on Chemistry Textbooks

Organic Chemistry

Read Explanation

Inorganic Chemistry

Read Explanation

Chemistry Branches

Read Explanation

Physical Chemistry

Read Explanation

Nuclear Chemistry

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.

Sign-up for free