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You have a 1.00-L sample of hot water (90.0°C) sitting open in a 25.0°C room. Eventually the water cools to 25.0°C while the temperature of the room remains unchanged. Calculate ∆S surr for this process. Assume the density of water is 1.00 g/cm 3 over this temperature range, and the heat capacity of water is constant over this temperature range and equal to 75.4 J/K mol.

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Chemistry: An Atoms First Approach

2nd Edition
Steven S. Zumdahl + 1 other
Publisher: Cengage Learning
ISBN: 9781305079243

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Chapter
Section
BuyFindarrow_forward

Chemistry: An Atoms First Approach

2nd Edition
Steven S. Zumdahl + 1 other
Publisher: Cengage Learning
ISBN: 9781305079243
Chapter 16, Problem 116CP
Textbook Problem
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You have a 1.00-L sample of hot water (90.0°C) sitting open in a 25.0°C room. Eventually the water cools to 25.0°C while the temperature of the room remains unchanged. Calculate ∆Ssurr for this process. Assume the density of water is 1.00 g/cm3 over this temperature range, and the heat capacity of water is constant over this temperature range and equal to 75.4 J/K mol.

Interpretation Introduction

Interpretation: The value of ΔSsurr for the given process is to be calculated.

Concept introduction: Thermodynamics is associated with heat, temperature and its relation with energy and work. It helps us to predict whether a process will take place or not. But it gives no information about the time required for the process. The terms associated with thermodynamics are system, surrounding, entropy, spontaneity and many more.

Explanation of Solution

Given

The density of water over the temperature range of 25 οC to 90.0 οC is 1.00g/cm3 .

The heat capacity of water in the given temperature range is 75.4J/Kmol .

The volume of water is 1L .

The conversion of L to mL is done as,

1L=1000mL

The conversion of mL to cm3 is done as,

1mL=1cm3

Therefore, the conversion of 1000mL to cm3 is done as,

1000mL=1000cm3

Therefore, the volume of water is 1000cm3 .

The mass of water is calculated by the formula,

Mass=Density×Volume

Substitute the value of density and volume in the above equation.

Mass=1.00g/cm3×1000cm3=1000g

Therefore, the mass of water is 1000g .

The number of moles of water is calculated by the formula,

Moles=MassMolarmass

The molar mass of water is 18g/mol .

Substitute the value of mass and molar mass of water in the above equation.

Moles=1000g18g/mol=55.55mol

The heat capacity for one mole of water is given as 75.4J/Kmol .

Therefore, the heat capacity of water for 55.55mol is 75.4J/Kmol×55.55mol=4188

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Chapter 16 Solutions

Chemistry: An Atoms First Approach
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