The specific heat of alloy has to be determined. Concept Introduction: Law of conservation of energy states that total energy of an isolated system remains constant, energy remains conserved overtime. This is first law of thermodynamics . Quantity of heat exchanged in an experiment is dependent on 3 factors-amounts of materials, specific heat capacity and the temperature change the material undergoes. The general relationship is as follows: heat = ( mass ) ( specific heat ) ( change in temeprature )

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Answer 1

Question

Chapter EN, Problem 1PQ

Interpretation Introduction

Interpretation:

The specific heat of alloy has to be determined.

Concept Introduction:

Law of conservation of energy states that total energy of an isolated system remains constant, energy remains conserved overtime. This is first law of thermodynamics.

Quantity of heat exchanged in an experiment is dependent on 3 factors-amounts of materials, specific heat capacity and the temperature change the material undergoes. The general relationship is as follows:

heat=(mass)(specific heat)(change in temeprature)

Expert Solution & Answer

Answer to Problem 1PQ

Option (B) is the correct one.

Explanation of Solution

When a sample of alloy is added to cooler water, heat flow will take place till thermal equilibrium is reached. Energy is conserved in this process, that is heat lost by silver is gained by water as per first law of thermodynamics.

The formula to calculate heat lost by alloy is a follows:

heat lost=(mass of alloy)(specific heat)(T2−T1) (1)

Here,

T2 is final temperature.

T1 is initial temperature.

Substitute 45 g for mass of alloy, 37 °C for T2 and 100 °C for T1 in equation (1).

heat lost=(45 g)(specific heat)(37 °C−100 °C)=−2835 g⋅°C(specific heat)

The formula to calculate heat gained by water is a follows:

heat gained=(mass of water)(specific heat)(T2−T1) (2)

Substitute 100 g for mass of alloy, 4.184 J⋅g−1⋅°C for specific heat, 37 °C for T2 and 25 °C for T1 in equation (2).

heat gained=(100 g)(4.184 J⋅g−1⋅°C−1)(37 °C−25 °C)=5020.8 J

Since energy remains conserved so it can be expressed as follows:

heat gained+heat lost=0

Substitute equation 5020.8 J for heat gained and −2835 g⋅°C(specific heat) for heat lost.

5020.8 J+(−2835 g⋅°C(specific heat))=0(specific heat)=−5020.8 J−2835 g⋅°C=1.771 J⋅g−1⋅°C−1

The specific heat of alloy is 1.771 J⋅g−1⋅°C−1. Hence option (B) is correct.

Reason for incorrect option:

Since specific heat of alloy is 1.771 J⋅g−1⋅°C−1. Hence, choices (A), (C) and (D) are incorrect.

Conclusion

The specific heat of alloy is 1.771 J⋅g−1⋅°C−1. Hence option (B) is correct.

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Answer 2

Question

Chapter EN, Problem 1PQ

Interpretation Introduction

Interpretation:

The specific heat of alloy has to be determined.

Concept Introduction:

Law of conservation of energy states that total energy of an isolated system remains constant, energy remains conserved overtime. This is first law of thermodynamics.

Quantity of heat exchanged in an experiment is dependent on 3 factors-amounts of materials, specific heat capacity and the temperature change the material undergoes. The general relationship is as follows:

heat=(mass)(specific heat)(change in temeprature)

Expert Solution & Answer

Answer to Problem 1PQ

Option (B) is the correct one.

Explanation of Solution

When a sample of alloy is added to cooler water, heat flow will take place till thermal equilibrium is reached. Energy is conserved in this process, that is heat lost by silver is gained by water as per first law of thermodynamics.

The formula to calculate heat lost by alloy is a follows:

heat lost=(mass of alloy)(specific heat)(T2−T1) (1)

Here,

T2 is final temperature.

T1 is initial temperature.

Substitute 45 g for mass of alloy, 37 °C for T2 and 100 °C for T1 in equation (1).

heat lost=(45 g)(specific heat)(37 °C−100 °C)=−2835 g⋅°C(specific heat)

The formula to calculate heat gained by water is a follows:

heat gained=(mass of water)(specific heat)(T2−T1) (2)

Substitute 100 g for mass of alloy, 4.184 J⋅g−1⋅°C for specific heat, 37 °C for T2 and 25 °C for T1 in equation (2).

heat gained=(100 g)(4.184 J⋅g−1⋅°C−1)(37 °C−25 °C)=5020.8 J

Since energy remains conserved so it can be expressed as follows:

heat gained+heat lost=0

Substitute equation 5020.8 J for heat gained and −2835 g⋅°C(specific heat) for heat lost.

5020.8 J+(−2835 g⋅°C(specific heat))=0(specific heat)=−5020.8 J−2835 g⋅°C=1.771 J⋅g−1⋅°C−1

The specific heat of alloy is 1.771 J⋅g−1⋅°C−1. Hence option (B) is correct.

Reason for incorrect option:

Since specific heat of alloy is 1.771 J⋅g−1⋅°C−1. Hence, choices (A), (C) and (D) are incorrect.

Conclusion

The specific heat of alloy is 1.771 J⋅g−1⋅°C−1. Hence option (B) is correct.

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Answer 3

Question

Chapter EN, Problem 1PQ

Interpretation Introduction

Interpretation:

The specific heat of alloy has to be determined.

Concept Introduction:

Law of conservation of energy states that total energy of an isolated system remains constant, energy remains conserved overtime. This is first law of thermodynamics.

Quantity of heat exchanged in an experiment is dependent on 3 factors-amounts of materials, specific heat capacity and the temperature change the material undergoes. The general relationship is as follows:

heat=(mass)(specific heat)(change in temeprature)

Expert Solution & Answer

Answer to Problem 1PQ

Option (B) is the correct one.

Explanation of Solution

When a sample of alloy is added to cooler water, heat flow will take place till thermal equilibrium is reached. Energy is conserved in this process, that is heat lost by silver is gained by water as per first law of thermodynamics.

The formula to calculate heat lost by alloy is a follows:

heat lost=(mass of alloy)(specific heat)(T2−T1) (1)

Here,

T2 is final temperature.

T1 is initial temperature.

Substitute 45 g for mass of alloy, 37 °C for T2 and 100 °C for T1 in equation (1).

heat lost=(45 g)(specific heat)(37 °C−100 °C)=−2835 g⋅°C(specific heat)

The formula to calculate heat gained by water is a follows:

heat gained=(mass of water)(specific heat)(T2−T1) (2)

Substitute 100 g for mass of alloy, 4.184 J⋅g−1⋅°C for specific heat, 37 °C for T2 and 25 °C for T1 in equation (2).

heat gained=(100 g)(4.184 J⋅g−1⋅°C−1)(37 °C−25 °C)=5020.8 J

Since energy remains conserved so it can be expressed as follows:

heat gained+heat lost=0

Substitute equation 5020.8 J for heat gained and −2835 g⋅°C(specific heat) for heat lost.

5020.8 J+(−2835 g⋅°C(specific heat))=0(specific heat)=−5020.8 J−2835 g⋅°C=1.771 J⋅g−1⋅°C−1

The specific heat of alloy is 1.771 J⋅g−1⋅°C−1. Hence option (B) is correct.

Reason for incorrect option:

Since specific heat of alloy is 1.771 J⋅g−1⋅°C−1. Hence, choices (A), (C) and (D) are incorrect.

Conclusion

The specific heat of alloy is 1.771 J⋅g−1⋅°C−1. Hence option (B) is correct.

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Answer 4

Question

Chapter EN, Problem 1PQ

Interpretation Introduction

Interpretation:

The specific heat of alloy has to be determined.

Concept Introduction:

Law of conservation of energy states that total energy of an isolated system remains constant, energy remains conserved overtime. This is first law of thermodynamics.

Quantity of heat exchanged in an experiment is dependent on 3 factors-amounts of materials, specific heat capacity and the temperature change the material undergoes. The general relationship is as follows:

heat=(mass)(specific heat)(change in temeprature)

Expert Solution & Answer

Answer to Problem 1PQ

Option (B) is the correct one.

Explanation of Solution

When a sample of alloy is added to cooler water, heat flow will take place till thermal equilibrium is reached. Energy is conserved in this process, that is heat lost by silver is gained by water as per first law of thermodynamics.

The formula to calculate heat lost by alloy is a follows:

heat lost=(mass of alloy)(specific heat)(T2−T1) (1)

Here,

T2 is final temperature.

T1 is initial temperature.

Substitute 45 g for mass of alloy, 37 °C for T2 and 100 °C for T1 in equation (1).

heat lost=(45 g)(specific heat)(37 °C−100 °C)=−2835 g⋅°C(specific heat)

The formula to calculate heat gained by water is a follows:

heat gained=(mass of water)(specific heat)(T2−T1) (2)

Substitute 100 g for mass of alloy, 4.184 J⋅g−1⋅°C for specific heat, 37 °C for T2 and 25 °C for T1 in equation (2).

heat gained=(100 g)(4.184 J⋅g−1⋅°C−1)(37 °C−25 °C)=5020.8 J

Since energy remains conserved so it can be expressed as follows:

heat gained+heat lost=0

Substitute equation 5020.8 J for heat gained and −2835 g⋅°C(specific heat) for heat lost.

5020.8 J+(−2835 g⋅°C(specific heat))=0(specific heat)=−5020.8 J−2835 g⋅°C=1.771 J⋅g−1⋅°C−1

The specific heat of alloy is 1.771 J⋅g−1⋅°C−1. Hence option (B) is correct.

Reason for incorrect option:

Since specific heat of alloy is 1.771 J⋅g−1⋅°C−1. Hence, choices (A), (C) and (D) are incorrect.

Conclusion

The specific heat of alloy is 1.771 J⋅g−1⋅°C−1. Hence option (B) is correct.

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Answer 5

Chapter EN, Problem 1PQ

Interpretation Introduction

Interpretation:

The specific heat of alloy has to be determined.

Concept Introduction:

Law of conservation of energy states that total energy of an isolated system remains constant, energy remains conserved overtime. This is first law of thermodynamics.

Quantity of heat exchanged in an experiment is dependent on 3 factors-amounts of materials, specific heat capacity and the temperature change the material undergoes. The general relationship is as follows:

heat=(mass)(specific heat)(change in temeprature)

Expert Solution & Answer

Answer to Problem 1PQ

Option (B) is the correct one.

Explanation of Solution

When a sample of alloy is added to cooler water, heat flow will take place till thermal equilibrium is reached. Energy is conserved in this process, that is heat lost by silver is gained by water as per first law of thermodynamics.

The formula to calculate heat lost by alloy is a follows:

heat lost=(mass of alloy)(specific heat)(T2−T1) (1)

Here,

T2 is final temperature.

T1 is initial temperature.

Substitute 45 g for mass of alloy, 37 °C for T2 and 100 °C for T1 in equation (1).

heat lost=(45 g)(specific heat)(37 °C−100 °C)=−2835 g⋅°C(specific heat)

The formula to calculate heat gained by water is a follows:

heat gained=(mass of water)(specific heat)(T2−T1) (2)

Substitute 100 g for mass of alloy, 4.184 J⋅g−1⋅°C for specific heat, 37 °C for T2 and 25 °C for T1 in equation (2).

heat gained=(100 g)(4.184 J⋅g−1⋅°C−1)(37 °C−25 °C)=5020.8 J

Since energy remains conserved so it can be expressed as follows:

heat gained+heat lost=0

Substitute equation 5020.8 J for heat gained and −2835 g⋅°C(specific heat) for heat lost.

5020.8 J+(−2835 g⋅°C(specific heat))=0(specific heat)=−5020.8 J−2835 g⋅°C=1.771 J⋅g−1⋅°C−1

The specific heat of alloy is 1.771 J⋅g−1⋅°C−1. Hence option (B) is correct.

Reason for incorrect option:

Since specific heat of alloy is 1.771 J⋅g−1⋅°C−1. Hence, choices (A), (C) and (D) are incorrect.

Conclusion

The specific heat of alloy is 1.771 J⋅g−1⋅°C−1. Hence option (B) is correct.

Answer 6

Chapter EN, Problem 1PQ

Interpretation Introduction

Interpretation:

The specific heat of alloy has to be determined.

Concept Introduction:

Law of conservation of energy states that total energy of an isolated system remains constant, energy remains conserved overtime. This is first law of thermodynamics.

Quantity of heat exchanged in an experiment is dependent on 3 factors-amounts of materials, specific heat capacity and the temperature change the material undergoes. The general relationship is as follows:

heat=(mass)(specific heat)(change in temeprature)

Expert Solution & Answer

Answer to Problem 1PQ

Option (B) is the correct one.

Explanation of Solution

When a sample of alloy is added to cooler water, heat flow will take place till thermal equilibrium is reached. Energy is conserved in this process, that is heat lost by silver is gained by water as per first law of thermodynamics.

The formula to calculate heat lost by alloy is a follows:

heat lost=(mass of alloy)(specific heat)(T2−T1) (1)

Here,

T2 is final temperature.

T1 is initial temperature.

Substitute 45 g for mass of alloy, 37 °C for T2 and 100 °C for T1 in equation (1).

heat lost=(45 g)(specific heat)(37 °C−100 °C)=−2835 g⋅°C(specific heat)

The formula to calculate heat gained by water is a follows:

heat gained=(mass of water)(specific heat)(T2−T1) (2)

Substitute 100 g for mass of alloy, 4.184 J⋅g−1⋅°C for specific heat, 37 °C for T2 and 25 °C for T1 in equation (2).

heat gained=(100 g)(4.184 J⋅g−1⋅°C−1)(37 °C−25 °C)=5020.8 J

Since energy remains conserved so it can be expressed as follows:

heat gained+heat lost=0

Substitute equation 5020.8 J for heat gained and −2835 g⋅°C(specific heat) for heat lost.

5020.8 J+(−2835 g⋅°C(specific heat))=0(specific heat)=−5020.8 J−2835 g⋅°C=1.771 J⋅g−1⋅°C−1

The specific heat of alloy is 1.771 J⋅g−1⋅°C−1. Hence option (B) is correct.

Reason for incorrect option:

Since specific heat of alloy is 1.771 J⋅g−1⋅°C−1. Hence, choices (A), (C) and (D) are incorrect.

Conclusion

The specific heat of alloy is 1.771 J⋅g−1⋅°C−1. Hence option (B) is correct.

Answer 7

Chapter EN, Problem 1PQ

Interpretation Introduction

Interpretation:

The specific heat of alloy has to be determined.

Concept Introduction:

Law of conservation of energy states that total energy of an isolated system remains constant, energy remains conserved overtime. This is first law of thermodynamics.

Quantity of heat exchanged in an experiment is dependent on 3 factors-amounts of materials, specific heat capacity and the temperature change the material undergoes. The general relationship is as follows:

heat=(mass)(specific heat)(change in temeprature)

Answer 8

Expert Solution & Answer

Answer to Problem 1PQ

Option (B) is the correct one.

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

Explanation of Solution

When a sample of alloy is added to cooler water, heat flow will take place till thermal equilibrium is reached. Energy is conserved in this process, that is heat lost by silver is gained by water as per first law of thermodynamics.

The formula to calculate heat lost by alloy is a follows:

heat lost=(mass of alloy)(specific heat)(T2−T1) (1)