The total amount of heat needed to change 0.500 kg of ice at − 10. ° C into 0.500 kg 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.

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Answer 1

Question

Definition Definition Substance that constitutes everything in the universe. Matter consists of atoms, which are composed of electrons, protons, and neutrons. Different atoms combine together to give rise to molecules that act as a foundation for all kinds of substances. There are five states of matter based on their energies of attraction: solid, liquid, gases, plasma, and BEC (Bose-Einstein condensates).

Chapter 6, Problem 6.81E

Interpretation Introduction

Interpretation:

The total amount of heat needed to change 0.500 kg of ice at −10.°C into 0.500 kg 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

Answer to Problem 6.81E

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

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=(sample mass)(specific heat)(temperature change)

Substitute the values in the above equation as follows.

Heat=(sample mass)(specific heat)(temperature change)=(0.500×1000g)(0.51 calg °C)(0°C−(−10°C))=(500 g)(0.51 calg °C)(10°C)=2550 cal

Step 2. Ice(0.°C)→water(0.°C)

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

Heat = (Heat of fusion)(sample mass)

Substitute the values in the above equation as follows.

Heat = (Heat of fusion)(sample mass)=(80 cal/g)(0.500 ×1000 g)=40000 cal

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=(sample mass)(specific heat)(temperature change)

Substitute the values in the above equation as follows.

Heat=(sample mass)(specific heat)(temperature change)=(0.500×1000g)(1.0 calg °C)(100°C−0°C)=(500 g)(1.0 calg °C)(100°C)=50000 cal

Step 4. Water(100.°C)→steam(100.°C)

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

Heat = (Heat of vaporization)(sample mass)

Substitute the values in the above equation as follows.

Heat = (Heat of vaporization)(sample mass)=(540 cal/g)(0.500 ×1000 g)=270000 cal

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=(sample mass)(specific heat)(temperature change)

Substitute the values in the above equation as follows.

Heat=(sample mass)(specific heat)(temperature change)=(0.500×1000g)(0.48 calg °C)(120°C−100°C)=(500 g)(0.48 calg °C)(20°C)=4800 cal

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

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

Conclusion

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

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Answer 2

Question

Definition Definition Substance that constitutes everything in the universe. Matter consists of atoms, which are composed of electrons, protons, and neutrons. Different atoms combine together to give rise to molecules that act as a foundation for all kinds of substances. There are five states of matter based on their energies of attraction: solid, liquid, gases, plasma, and BEC (Bose-Einstein condensates).

Chapter 6, Problem 6.81E

Interpretation Introduction

Interpretation:

The total amount of heat needed to change 0.500 kg of ice at −10.°C into 0.500 kg 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

Answer to Problem 6.81E

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

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=(sample mass)(specific heat)(temperature change)

Substitute the values in the above equation as follows.

Heat=(sample mass)(specific heat)(temperature change)=(0.500×1000g)(0.51 calg °C)(0°C−(−10°C))=(500 g)(0.51 calg °C)(10°C)=2550 cal

Step 2. Ice(0.°C)→water(0.°C)

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

Heat = (Heat of fusion)(sample mass)

Substitute the values in the above equation as follows.

Heat = (Heat of fusion)(sample mass)=(80 cal/g)(0.500 ×1000 g)=40000 cal

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=(sample mass)(specific heat)(temperature change)

Substitute the values in the above equation as follows.

Heat=(sample mass)(specific heat)(temperature change)=(0.500×1000g)(1.0 calg °C)(100°C−0°C)=(500 g)(1.0 calg °C)(100°C)=50000 cal

Step 4. Water(100.°C)→steam(100.°C)

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

Heat = (Heat of vaporization)(sample mass)

Substitute the values in the above equation as follows.

Heat = (Heat of vaporization)(sample mass)=(540 cal/g)(0.500 ×1000 g)=270000 cal

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=(sample mass)(specific heat)(temperature change)

Substitute the values in the above equation as follows.

Heat=(sample mass)(specific heat)(temperature change)=(0.500×1000g)(0.48 calg °C)(120°C−100°C)=(500 g)(0.48 calg °C)(20°C)=4800 cal

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

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

Conclusion

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

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Answer 3

Question

Definition Definition Substance that constitutes everything in the universe. Matter consists of atoms, which are composed of electrons, protons, and neutrons. Different atoms combine together to give rise to molecules that act as a foundation for all kinds of substances. There are five states of matter based on their energies of attraction: solid, liquid, gases, plasma, and BEC (Bose-Einstein condensates).

Chapter 6, Problem 6.81E

Interpretation Introduction

Interpretation:

The total amount of heat needed to change 0.500 kg of ice at −10.°C into 0.500 kg 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

Answer to Problem 6.81E

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

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=(sample mass)(specific heat)(temperature change)

Substitute the values in the above equation as follows.

Heat=(sample mass)(specific heat)(temperature change)=(0.500×1000g)(0.51 calg °C)(0°C−(−10°C))=(500 g)(0.51 calg °C)(10°C)=2550 cal

Step 2. Ice(0.°C)→water(0.°C)

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

Heat = (Heat of fusion)(sample mass)

Substitute the values in the above equation as follows.

Heat = (Heat of fusion)(sample mass)=(80 cal/g)(0.500 ×1000 g)=40000 cal

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=(sample mass)(specific heat)(temperature change)

Substitute the values in the above equation as follows.

Heat=(sample mass)(specific heat)(temperature change)=(0.500×1000g)(1.0 calg °C)(100°C−0°C)=(500 g)(1.0 calg °C)(100°C)=50000 cal

Step 4. Water(100.°C)→steam(100.°C)

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

Heat = (Heat of vaporization)(sample mass)

Substitute the values in the above equation as follows.

Heat = (Heat of vaporization)(sample mass)=(540 cal/g)(0.500 ×1000 g)=270000 cal

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=(sample mass)(specific heat)(temperature change)

Substitute the values in the above equation as follows.

Heat=(sample mass)(specific heat)(temperature change)=(0.500×1000g)(0.48 calg °C)(120°C−100°C)=(500 g)(0.48 calg °C)(20°C)=4800 cal

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

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

Conclusion

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

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Answer 4

Question

Definition Definition Substance that constitutes everything in the universe. Matter consists of atoms, which are composed of electrons, protons, and neutrons. Different atoms combine together to give rise to molecules that act as a foundation for all kinds of substances. There are five states of matter based on their energies of attraction: solid, liquid, gases, plasma, and BEC (Bose-Einstein condensates).

Chapter 6, Problem 6.81E

Interpretation Introduction

Interpretation:

The total amount of heat needed to change 0.500 kg of ice at −10.°C into 0.500 kg 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

Answer to Problem 6.81E

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

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=(sample mass)(specific heat)(temperature change)

Substitute the values in the above equation as follows.

Heat=(sample mass)(specific heat)(temperature change)=(0.500×1000g)(0.51 calg °C)(0°C−(−10°C))=(500 g)(0.51 calg °C)(10°C)=2550 cal

Step 2. Ice(0.°C)→water(0.°C)

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

Heat = (Heat of fusion)(sample mass)

Substitute the values in the above equation as follows.

Heat = (Heat of fusion)(sample mass)=(80 cal/g)(0.500 ×1000 g)=40000 cal

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=(sample mass)(specific heat)(temperature change)

Substitute the values in the above equation as follows.

Heat=(sample mass)(specific heat)(temperature change)=(0.500×1000g)(1.0 calg °C)(100°C−0°C)=(500 g)(1.0 calg °C)(100°C)=50000 cal

Step 4. Water(100.°C)→steam(100.°C)

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

Heat = (Heat of vaporization)(sample mass)

Substitute the values in the above equation as follows.

Heat = (Heat of vaporization)(sample mass)=(540 cal/g)(0.500 ×1000 g)=270000 cal

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=(sample mass)(specific heat)(temperature change)

Substitute the values in the above equation as follows.

Heat=(sample mass)(specific heat)(temperature change)=(0.500×1000g)(0.48 calg °C)(120°C−100°C)=(500 g)(0.48 calg °C)(20°C)=4800 cal

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

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

Conclusion

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

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Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 5

Chapter 6, Problem 6.81E

Interpretation Introduction

Interpretation:

The total amount of heat needed to change 0.500 kg of ice at −10.°C into 0.500 kg 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

Answer to Problem 6.81E

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

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=(sample mass)(specific heat)(temperature change)

Substitute the values in the above equation as follows.

Heat=(sample mass)(specific heat)(temperature change)=(0.500×1000g)(0.51 calg °C)(0°C−(−10°C))=(500 g)(0.51 calg °C)(10°C)=2550 cal

Step 2. Ice(0.°C)→water(0.°C)

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

Heat = (Heat of fusion)(sample mass)

Substitute the values in the above equation as follows.

Heat = (Heat of fusion)(sample mass)=(80 cal/g)(0.500 ×1000 g)=40000 cal

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=(sample mass)(specific heat)(temperature change)

Substitute the values in the above equation as follows.

Heat=(sample mass)(specific heat)(temperature change)=(0.500×1000g)(1.0 calg °C)(100°C−0°C)=(500 g)(1.0 calg °C)(100°C)=50000 cal

Step 4. Water(100.°C)→steam(100.°C)

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

Heat = (Heat of vaporization)(sample mass)

Substitute the values in the above equation as follows.

Heat = (Heat of vaporization)(sample mass)=(540 cal/g)(0.500 ×1000 g)=270000 cal

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=(sample mass)(specific heat)(temperature change)

Substitute the values in the above equation as follows.

Heat=(sample mass)(specific heat)(temperature change)=(0.500×1000g)(0.48 calg °C)(120°C−100°C)=(500 g)(0.48 calg °C)(20°C)=4800 cal

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

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

Conclusion

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

Answer 6

Chapter 6, Problem 6.81E

Interpretation Introduction

Interpretation:

The total amount of heat needed to change 0.500 kg of ice at −10.°C into 0.500 kg 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

Answer to Problem 6.81E

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

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=(sample mass)(specific heat)(temperature change)

Substitute the values in the above equation as follows.

Heat=(sample mass)(specific heat)(temperature change)=(0.500×1000g)(0.51 calg °C)(0°C−(−10°C))=(500 g)(0.51 calg °C)(10°C)=2550 cal

Step 2. Ice(0.°C)→water(0.°C)

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

Heat = (Heat of fusion)(sample mass)

Substitute the values in the above equation as follows.

Heat = (Heat of fusion)(sample mass)=(80 cal/g)(0.500 ×1000 g)=40000 cal

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=(sample mass)(specific heat)(temperature change)

Substitute the values in the above equation as follows.

Heat=(sample mass)(specific heat)(temperature change)=(0.500×1000g)(1.0 calg °C)(100°C−0°C)=(500 g)(1.0 calg °C)(100°C)=50000 cal

Step 4. Water(100.°C)→steam(100.°C)

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

Heat = (Heat of vaporization)(sample mass)

Substitute the values in the above equation as follows.

Heat = (Heat of vaporization)(sample mass)=(540 cal/g)(0.500 ×1000 g)=270000 cal

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=(sample mass)(specific heat)(temperature change)

Substitute the values in the above equation as follows.

Heat=(sample mass)(specific heat)(temperature change)=(0.500×1000g)(0.48 calg °C)(120°C−100°C)=(500 g)(0.48 calg °C)(20°C)=4800 cal

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

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

Conclusion

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

Answer 7

Chapter 6, Problem 6.81E

Interpretation Introduction

Interpretation:

The total amount of heat needed to change 0.500 kg of ice at −10.°C into 0.500 kg 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.

Answer 8

Expert Solution & Answer

Answer to Problem 6.81E

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

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

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=(sample mass)(specific heat)(temperature change)

Substitute the values in the above equation as follows.

Heat=(sample mass)(specific heat)(temperature change)=(0.500×1000g)(0.51 calg °C)(0°C−(−10°C))=(500 g)(0.51 calg °C)(10°C)=2550 cal

Step 2. Ice(0.°C)→water(0.°C)

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

Heat = (Heat of fusion)(sample mass)

Substitute the values in the above equation as follows.

Heat = (Heat of fusion)(sample mass)=(80 cal/g)(0.500 ×1000 g)=40000 cal

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=(sample mass)(specific heat)(temperature change)

Substitute the values in the above equation as follows.

Heat=(sample mass)(specific heat)(temperature change)=(0.500×1000g)(1.0 calg °C)(100°C−0°C)=(500 g)(1.0 calg °C)(100°C)=50000 cal

Step 4. Water(100.°C)→steam(100.°C)

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

Heat = (Heat of vaporization)(sample mass)

Substitute the values in the above equation as follows.

Heat = (Heat of vaporization)(sample mass)=(540 cal/g)(0.500 ×1000 g)=270000 cal

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=(sample mass)(specific heat)(temperature change)

Substitute the values in the above equation as follows.

Heat=(sample mass)(specific heat)(temperature change)=(0.500×1000g)(0.48 calg °C)(120°C−100°C)=(500 g)(0.48 calg °C)(20°C)=4800 cal

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

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

Answer 12

Conclusion

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

Answer 13

Ch. 6 - Calculate the volume of 125g of the following...Ch. 6 - Calculate the volume of 125g of the following...Ch. 6 - Copper metal has a density of 8.92g/cm3 at 20.0C...Ch. 6 - Liquid water has a density of 1.00g/mL at 10.0C...Ch. 6 - Gallium metal melts at 29.8C. At the melting...Ch. 6 - Prob. 6.6E Ch. 6 - Describe the change in form of energy kinetic...Ch. 6 - Prob. 6.8E Ch. 6 - Prob. 6.9E Ch. 6 - At 25.0C, helium molecules (He) have an average...

Answer to Problem 6.81E

Explanation of Solution

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Chapter 6 Solutions