Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 4: Problem 4

The heat capacity of liquid water is \(75.6 \mathrm{~J} / \mathrm{K}-\mathrm{mol}\), while the enthalpy of fusion of ice is \(6.0 \mathrm{~kJ} / \mathrm{mol}\). What is the smallest number of ice cubes at \(0^{\circ} \mathrm{C}\), each containing \(9.0 \mathrm{~g}\) of water, needed to cool \(500 \mathrm{~g}\) of liquid water from \(20^{\circ} \mathrm{C}\) to \(0^{\circ} \mathrm{C}\) ? (a) 1 (b) 7 (c) 14 (d) 21

Short Answer

Expert verified
(b) 7

Step by step solution

01

Calculate the heat removed from liquid water

First, calculate how much heat is removed from the 500 g of liquid water when it cools from 20°C to 0°C. Use the formula: Heat removed (Q) = mass (m) * specific heat capacity (C) * change in temperature (ΔT). Note that the heat capacity of the water is given per mole, so we will need to convert the mass of the water to moles before using it. The molar mass of water is 18.015 g/mol.
02

Convert mass of water to moles

Use the molar mass of water (18.015 g/mol) to convert 500 g of liquid water to moles: number of moles (n) = mass (m) / molar mass (M).
03

Calculate the heat removed from the liquid water

Plug in the values for number of moles, the specific heat capacity, and the change in temperature to get the total heat removed.
04

Calculate enthalpy change for the fusion of ice

Use the enthalpy of fusion of ice to find out how much heat is needed to melt each ice cube. The enthalpy of fusion is given for 1 mole of ice, so there is no need to convert the mass of ice to moles as it will cancel out in the end.
05

Calculate the heat absorbed by melting ice cubes

Divide the total heat removed from liquid water by the heat absorbed per ice cube to find out how many ice cubes are needed.
06

Provide the correct answer option

Round up the number of ice cubes needed because you cannot use a fraction of an ice cube.

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!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Enthalpy of Fusion
The enthalpy of fusion is a measure of the heat energy required to change a substance from solid to liquid at constant pressure without changing its temperature. For ice, this process involves absorbing heat as the ice melts into liquid water. In our exercise, the enthalpy of fusion is crucial for determining how much energy each ice cube can absorb as it melts and subsequently cools the surrounding water.
Heat Capacity
Heat capacity represents the amount of heat needed to raise the temperature of a substance by one degree Celsius. It's a property that can help us understand how a substance responds to heat. In the context of our problem, the heat capacity of liquid water tells us how much energy is required to lower the temperature of a specific quantity of water from one temperature to another.
Calorimetry
Calorimetry is the science of measuring the change in heat associated with a chemical reaction or physical process. In this scenario, calorimetry principles are applied to calculate the energy exchange that occurs when the water temperature decreases, and ice melts. In simple terms, we're using calorimetry to balance the heat lost by the water and the heat gained by the ice.
Thermochemistry
Thermochemistry is a branch of chemistry concerned with the quantities of heat evolved or absorbed during chemical reactions. It deals with the energy changes in reactions, including fusion and phase changes. For the given problem, thermochemistry principles guide us in understanding how heat transfer between ice and water leads to cooling the water down to a desired temperature.
Molar Mass Calculation
Molar mass calculation is important in converting between mass and moles, as these conversions are often necessary in stoichiometry problems and thermochemical calculations. Our problem involves such a conversion to determine the number of moles of water being cooled. Given the molar mass of water and the mass of the water to be cooled, we can calculate the moles and use this amount in subsequent heat calculations.

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 quantity of \(1.6 \mathrm{~g}\) helium gas is expanded adiabatically \(3.0\) times and then compressed isobarically to the initial volume. Assume ideal behaviour of gas and both the processes reversible. The entropy change of the gas in this process is \((\ln 3=1.1)\) (a) \(-1.1 \mathrm{cal} / \mathrm{K}\) (b) \(+1.1 \mathrm{cal} / \mathrm{K}\) (c) \(-2.2 \mathrm{cal} / \mathrm{K}\) (d) \(+2.2 \mathrm{cal} / \mathrm{K}\)

In the reversible adiabatic expansion of an ideal monoatomic gas, the final volume is 8 times the initial volume. The ratio of final temperature to initial temperature is (a) \(8: 1\) (b) \(1: 4\) (c) \(1: 2\) (d) \(4: 1\)

Two moles of an ideal monoatomic gas is heated from \(27^{\circ} \mathrm{C}\) to \(627^{\circ} \mathrm{C}\), reversibly and isochorically. The entropy of gas (a) increases by \(2 R \ln 3\) (b) increases by \(3 R \ln 3\) (c) decreases by \(2 R \ln 3\) (d) decreases by \(3 R \ln 3\)

A diatomic ideal gas initially at \(273 \mathrm{~K}\) is given 100 cal heat due to which system did \(210 \mathrm{~J}\) work. Molar heat capacity of the gas for the process is \((1 \mathrm{cal}=4.2 \mathrm{~J})\) (a) \(\frac{3}{2} R\) (b) \(\frac{5}{2} R\) (c) \(\frac{5}{4} R\) (d) \(5 R\)

For a reaction: \(\mathrm{A} \rightleftharpoons \mathrm{B}\), carried out at \(27^{\circ} \mathrm{C}\), the ratio of equilibrium concentrations of product to reactant changes by a factor of \(\mathrm{e}^{4}\) for every (a) \(1.2\) kcal rise in \(\Delta G^{\circ}\) (b) \(1.2 \mathrm{kcal}\) fall in \(\Delta G^{\circ}\) (c) \(2.4 \mathrm{kcal}\) rise in \(\Delta G^{\circ}\) (d) \(2.4\) kcal fall in \(\Delta G^{\circ}\)
See all solutions

Recommended explanations on Chemistry Textbooks

Organic Chemistry

Read Explanation

Kinetics

Read Explanation

Inorganic Chemistry

Read Explanation

Chemical Reactions

Read Explanation

Chemical Analysis

Read Explanation

The Earths Atmosphere

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