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Introductory Chemistry: A Foundati...

9th Edition
Steven S. Zumdahl + 1 other
ISBN: 9781337399425

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BuyFindarrow_forward

Introductory Chemistry: A Foundati...

9th Edition
Steven S. Zumdahl + 1 other
ISBN: 9781337399425
Textbook Problem
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Consider four 100.0-g samples of water, each in a separate beaker at 25.0 °C. Into each beaker you drop 10.0 g of a different metal that has been healed to 95.0 °C. Assuming no heat loss to the surroundings, which water sample will have the highest final temperature? Explain your answer.

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  • The water to which you have added aluminum Chapter 10, Problem 16ALQ, Consider four 100.0-g samples of water, each in a separate beaker at 25.0 °C. Into each beaker you , example  1°C).
  • i>The water to which you have added iron Chapter 10, Problem 16ALQ, Consider four 100.0-g samples of water, each in a separate beaker at 25.0 °C. Into each beaker you , example  2°C)

    i>The water to which you have added copper Chapter 10, Problem 16ALQ, Consider four 100.0-g samples of water, each in a separate beaker at 25.0 °C. Into each beaker you , example  3°C).

    i>The water to which you have added lead Chapter 10, Problem 16ALQ, Consider four 100.0-g samples of water, each in a separate beaker at 25.0 °C. Into each beaker you , example  4°C).

    i>Because the masses of the metals are the same, the final temperatures would be the same.

    Interpretation Introduction

    (a)

    Interpretation:

    In the given reaction, which water sample will have the highest final temperature should be determined.

    Concept Introduction:

    The Calorimetric expression is

    Q=ms×DTs×SHs----(1)Here, Q is the heat involved, msis the mass of the sampleDTsis change in temperature and SHs is the specific heat of sample.As given ms=100.0gWe know, DTs=T2-T1T2 is the final temperature and T1=25oCSubstituting these values in the equation 1Qw=418.4×(T2-25)J/oC.

    Explanation

    Metal given is aluminium, mass of aluminium is 10.0 g, SHs for aluminium is 0.89J/g0 C. Assume the final temperature of the sample be T2 andT1 = 95 0 C

    Substitute these values in the equation 1

    QAl=8.9(T2-95)J/goC

    We know that the heat lost by metal is same as heat gained by water, so we have

    -QAl=Qw-8

    Interpretation Introduction

    (b)

    Interpretation:

    In the given reaction, which water sample will have the highest final temperature should be determined.

    Concept Introduction:

    The Calorimetric expression is

    Q=ms×DTs×SHs----(1)Here, Q is the heat involved, msis the mass of the sampleDTsis change in temperature and SHs is the specific heat of sample.As given ms=100.0gWe know, DTs=T2-T1T2 is the final temperature and T1=25oCSubstituting these values in the equation 1Qw=418.4×(T2-25)J/oC.

    Interpretation Introduction

    (c)

    Interpretation:

    In the given reaction, which water sample will have the highest final temperature should be determined.

    Concept Introduction:

    The Calorimetric expression is

    Q=ms×ΔTs×SHs----(1)Here, Q is the heat involved,  msis the mass of the sampleΔTsis change in temperature and SHs is the specific heat of sample.As given  ms=100.0gWe know,  ΔTs=T2-T1T2 is the final temperature and T1=25oCSubstituting these values in the equation 1Qw=418.4×(T2-25)J/oC.

    Interpretation Introduction

    (d)

    Interpretation:

    In the given reaction, which water sample will have the highest final temperature should be determined.

    Concept Introduction:

    The Calorimetric expression is

    Q=ms×DTs×SHs----(1)Here, Q is the heat involved, msis the mass of the sampleDTsis change in temperature and SHs is the specific heat of sample.As given ms=100.0gWe know, DTs=T2-T1T2 is the final temperature and T1=25oCSubstituting these values in the equation 1Qw=418.4×(T2-25)J/oC.

    Interpretation Introduction

    (e)

    Interpretation:

    In the given problem if the mass of all the samples is the same then final temperatures would same or not should be determined.

    Concept Introduction:

    The Calorimetric expression is

    Q=ms×ΔTs×SHs----(1)Here, Q is the heat involved,  msis the mass of the sampleΔTsis change in temperature and SHs is the specific heat of sample.As given  ms=100.0gWe know,  ΔTs=T2-T1T2 is the final temperature and T1=25oCSubstituting these values in the equation 1Qw=418.4×(T2-25)J/oC.

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