Integrated Science
Integrated Science
7th Edition
ISBN: 9780077862602
Author: Tillery, Bill W.
Publisher: Mcgraw-hill,
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Chapter 4, Problem 11PEB
To determine

The amount of heat removed by the cooling tower for each kilogram of spent steam.

Expert Solution & Answer
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Answer to Problem 11PEB

The amount of heat removed by the cooling tower for each kilogram of spent steam is 560kcal.

Explanation of Solution

To change the steam at 120.0°C into water at 90.0°C, three different quantities of heat are required.

First is the heat required to lower the temperature of steam to 100.0°C. Second is the heat required to cause a phase change from steam to water at 100.0°C. Third is the heat required to lower the temperature of water to 90.0°C.

Write the expression for the total heat removed.

    Q=Q1+Q2+Q3                                                               (I)

Here, Q is the total heat energy removed, Q1 is the heat removed to lower the temperature of steam, Q2 is the heat required for phase change of steam and Q3 is the heat removed to lower the temperature of water.

Write the expression for the heat removed to lower the temperature of steam.

    Q1=mcsteamΔTsteam                                                           (II)

Here, m is the mass of the steam, csteam is the specific heat of steam and ΔTsteam is the change in temperature of steam.

Steam completely changes to water. Hence, the mass of water is same as the mass of steam.

Write the expression for the heat removed for phase change of steam.

    Q2=mLv                                                                       (III)

Here, Q2 is the heat removed for phase change of steam, m is the mass of steam and Lv is the latent heat of vaporization for water.

Write the expression for the heat removed to lower the temperature of water.

    Q3=mcwaterΔTwater                                                         (IV)

Here, m is the mass of the water, cwater is the specific heat of water and ΔTwater is the change in temperature of water.

Write the expression for the change in temperature.

    ΔT=T1T2                                                               (V)

Here, T2 is the final temperature and T1 is the initial temperature.

Conclusion:

Substitute 100.0°C for T1 and 120.0°C for T2 in equation (V) to find ΔTsteam.

    ΔTsteam=120.0°C100.0°C=20.0°C

Substitute 20.0°C for ΔTsteam and 1.0kg for m and 0.480cal/g°C for csteam in equation (II) to find Q1.

    Q1=(1.0kg×103g1kg)(0.480cal/g°C)(20.0°C)=9600cal

Substitute 1.0kg for m and 540.0cal/g for Lv in equation (III) to find Q2.

    Q2=(1.0kg×103g1kg)(540.0cal/g)=540000cal

Substitute 90.0°C for T2 and 100.0°C for T1 in equation (V) to find ΔTwater.

    ΔTwater=100.0°C90.0°C=10.0°C

Substitute 10.0°C for ΔTwater, 1.0kg for m and 1.00cal/g°C for cwater in equation (IV) to find Q3.

    Q3=(1.0kg×103g1kg)(1.00cal/g°C)(10.0°C)=10000cal

Substitute 9600cal for Q1, 540000cal for Q2 and 10000cal for Q3 in equation (I) to find Q.

    Q=9600cal+540000cal+10000cal=(559600cal×103kcal1cal)560kcal

Therefore, the amount of heat removed by the cooling tower for each kilogram of spent steam is 560kcal.

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