Integrated Science
Integrated Science
7th Edition
ISBN: 9780077862602
Author: Tillery, Bill W.
Publisher: Mcgraw-hill,
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Chapter 4, Problem 12PEA
To determine

The total heat that was removed from water.

Expert Solution & Answer
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Answer to Problem 12PEA

The total heat removed from water is 40.2kcal.

Explanation of Solution

Write the expression to calculate the change in temperature.

    ΔT=T1T2                                                    (I)

Here, ΔT is the change in temperature, T1 is initial temperature and T2 is the final temperature.

Write the expression to calculate the heat released.

    Q1=mcΔT                                                   (II)

Here, Q1 is the heat released from the water, m is the mass, c is the specific heat and ΔT is the change in temperature.

Write the expression to calculate the heat released for the phase change.

    Q2=mLf                                                    (III)

Here, Q2 is the heat released for the phase change, m is the mass and Lf is the latent heat of fusion.

Write the expression to calculate the total heat removed from water.

    Qtotal=Q1+Q2+Q3                                     (IV)

Here, Qtotal is the total heat removed from water and Q3 is the heat released from the ice when it cools from 0°C to 5.00°C.

Conclusion:

Case 1: The water cools from 18.0°C to 0°C.

Substitute 18.0°C for T1 and 0°C for T2 in equation (I) to find the change in temperature.

    ΔT=18.0°C0°C=18.0°C

Substitute 400.0g for m, 1.00cal/g°C for c and 18.0°C for ΔT in equation (II) to find the heat released from the water.

    Q1=(400.0g)(1.00cal/g°C)(18.0°C)=7,200cal(1kcal1000cal)=7.2kcal

Case 2: The phase changes from water at 0°C to ice at 0°C.

Substitute 400.0g for m and 80.0cal/g for Lf in equation (III) to find the heat released for the phase change.

    Q2=(400.0g)(80.0cal/g)=32,000cal(1kcal1000cal)=32.0kcal

Case 3: The ice cool from 0°C to 5.00°C.

Substitute 0°C for T1 and 5.00°C for T2 in equation (I) to find the change in temperature.

    ΔT=0°C(5.00°C)=5.00°C

Substitute 400.0g for m, 0.50cal/g°C for c and 5.00°C for ΔT in equation (II) to find the energy released from the ice.

    Q3=(400.0g)(0.50cal/g°C)(5.00°C)=1,000cal(1kcal1000cal)=1.0kcal

Substitute 7.2kcal for Q1, 32.0kcal for Q2 and 1.0kcal for Q3 in equation (IV) to find the total heat removed from water.

    Qtotal=7.2kcal+32.0kcal+1.0kcal=40.2kcal

Therefore, the total heat removed from water is 40.2kcal.

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