Chemistry for Today: General, Organic, and Biochemistry
Chemistry for Today: General, Organic, and Biochemistry
9th Edition
ISBN: 9781305960060
Author: Spencer L. Seager, Michael R. Slabaugh, Maren S. Hansen
Publisher: Cengage Learning
Question
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Chapter 6, Problem 6.81E
Interpretation Introduction

Interpretation:

The total amount of heat needed to change 0.500kg of ice at 10.°C into 0.500kg of steam at 120.°C is to be calculated.

Concept introduction:

When temperature is changed or the state of matter is changed the energy is either absorbed or released. The amount of energy needed to change temperature of matter is known as specific heat of matter. The amount of energy needed to change state of matter is known as heat fusion or vaporization.

Expert Solution & Answer
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Answer to Problem 6.81E

The total amount of heat needed to change 0.500kg of ice at 10.°C into 0.500kg of steam at 120.°C is 367350cal.

Explanation of Solution

Step 1. Ice(10.°C)Ice(0.°C)

The formula to calculate amount of heat (in calories) needed to increase the temperature is given below as,

Heat=(samplemass)(specificheat)(temperaturechange)

Substitute the values in the above equation as follows.

Heat=(samplemass)(specificheat)(temperaturechange)=(0.500×1000g)(0.51calg°C)(0°C(10°C))=(500g)(0.51calg°C)(10°C)=2550cal

Step 2. Ice(0.°C)water(0.°C)

The heat of fusion is defined as the energy required for changing the state of 1g of substance. Thus, the energy that could be stored by melting a solid is given by the formula written below.

Heat= (Heatof fusion)(samplemass)

Substitute the values in the above equation as follows.

Heat= (Heatof fusion)(samplemass)=(80cal/g)(0.500×1000g)=40000cal

Step 3. Water(0.°C)water(100.°C)

The formula to calculate amount of heat (in calories) needed to increase the temperature is given below as,

Heat=(samplemass)(specificheat)(temperaturechange)

Substitute the values in the above equation as follows.

Heat=(samplemass)(specificheat)(temperaturechange)=(0.500×1000g)(1.0calg°C)(100°C0°C)=(500g)(1.0calg°C)(100°C)=50000cal

Step 4. Water(100.°C)steam(100.°C)

The heat of vaporization is defined as the energy required for changing the state of 1g of substance. Thus, the energy that could be stored by melting a solid is given by the formula written below.

Heat= (Heatof vaporization)(samplemass)

Substitute the values in the above equation as follows.

Heat= (Heatof vaporization)(samplemass)=(540cal/g)(0.500×1000g)=270000cal

Step 5. Steam(100.°C)steam(120.°C)

The formula to calculate amount of heat (in calories) needed to increase the temperature is given below as,

Heat=(samplemass)(specificheat)(temperaturechange)

Substitute the values in the above equation as follows.

Heat=(samplemass)(specificheat)(temperaturechange)=(0.500×1000g)(0.48calg°C)(120°C100°C)=(500g)(0.48calg°C)(20°C)=4800cal

The total amount of heat needed to change 0.500kg of ice at 10.°C into 0.500kg of steam at 120.°C is calculated as,

Heat=2550cal+40000cal+50000cal+270000cal+4800cal=367350cal

Conclusion

The total amount of heat needed to change 0.500kg of ice at 10.°C into 0.500kg of steam at 120.°C is 367350cal.

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Chapter 6 Solutions

Chemistry for Today: General, Organic, and Biochemistry

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