Physics
Physics
3rd Edition
ISBN: 9780073512150
Author: Alan Giambattista, Betty Richardson, Robert C. Richardson Dr.
Publisher: McGraw-Hill Education
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Chapter 15, Problem 73P

(a)

To determine

The entropy change in the universe.

(a)

Expert Solution
Check Mark

Answer to Problem 73P

The change in entropy of universe is 0.90J/K.

Explanation of Solution

The mass of the first block of iron is 0.500kg and it is at temperature 60.0°C. The mass of the second block of iron is 0.500kg and it is at temperature 20.0°C. The common temperature after the blocks come into contact is 40.0°C.

Write the formula for the change in entropy of a system.

ΔS=QT

Here, ΔS is the change in entropy, Q is the heat flow into the system, T is the temperature.

The change in entropy of the universe is the sum of change of entropy of the two blocks.

ΔSU=ΔS1+ΔS2

Here, ΔSU is the change in entropy of the universe, ΔS1 is the change in entropy of the first block, ΔS2 is the change in entropy of the second bag.

Re-write the above equation by substituting the individual expressions.

ΔSU=Q1T1+Q2T2

Here, Q1 is the heat flow into the first block, Q2 is the heat flow into the second block, T1 is the average temperature of the first block, T2 is the average temperature of the second block.

Re-write the above equation by substituting the individual expressions.

ΔSU=m1cΔT1T1+m2cΔT2T2

Here, m1 is the mass of the first block, m2 is the mass of the second block, ΔT1 is the change in temperature of the first block, ΔT2 is the change in temperature of the second block, c is the specific heat capacity of the iron.

Conclusion:

Substitute 0.500kg for m1, 0.500kg for m2, 0.44×103J/kgK for c, 40.0°C-20.0°C for ΔT1, 40.0°C-60.0°C for ΔT2, (40.0°C+20.0°C)/2 for T1, (40.0°C+60.0°C)/2 for T2

ΔSU=((0.500kg)(0.44×103J/kgK))[(40.0°C-20.0°C)(40.0°C+20.0°C)/2+(40.0°C-60.0°C)(40.0°C+60.0°C)/2]=((0.500kg)(0.44×103J/kgK))[(40.0°C-20.0°C)30.0+273.15K+(40.0°C-60.0°C)50.0+273.15K]=0.90J/K

The change in entropy of universe is 0.90J/K.

(b)

To determine

The entropy change in the universe.

(b)

Expert Solution
Check Mark

Answer to Problem 73P

The change in entropy of universe is 2.7J/K.

Explanation of Solution

The mass of the first block of iron is 0.500kg and it is at temperature 60.0°C. The mass of the second block of iron is 0.500kg and it is at temperature 20.0°C. After the two blocks come into contact, the temperature of the first block increases to 80.0°C and the temperature of the second block decreases to 0.0°C.

Write the formula for the change in entropy of a system.

ΔS=QT

Here, ΔS is the change in entropy, Q is the heat flow into the system, T is the temperature.

The change in entropy of the universe is the sum of change of entropy of the two blocks.

ΔSU=ΔS1+ΔS2

Here, ΔSU is the change in entropy of the universe, ΔS1 is the change in entropy of the first block, ΔS2 is the change in entropy of the second bag.

Re-write the above equation by substituting the individual expressions.

ΔSU=Q1T1+Q2T2

Here, Q1 is the heat flow into the first block, Q2 is the heat flow into the second block, T1 is the average temperature of the first block, T2 is the average temperature of the second block.

Re-write the above equation by substituting the individual expressions.

ΔSU=m1cΔT1T1+m2cΔT2T2

Here, m1 is the mass of the first block, m2 is the mass of the second block, ΔT1 is the change in temperature of the first block, ΔT2 is the change in temperature of the second block, c is the specific heat capacity of the iron.

Conclusion:

Substitute 0.500kg for m1, 0.500kg for m2, 0.44×103J/kgK for c, 0.0°C-20.0°C for ΔT1, 80.0°C-60.0°C for ΔT2, (0.0°C+20.0°C)/2 for T1, (80.0°C+60.0°C)/2 for T2

ΔSU=((0.500kg)(0.44×103J/kgK))[(0.0°C-20.0°C)(0.0°C+20.0°C)/2+(80.0°C-60.0°C)(80.0°C+60.0°C)/2]=((0.500kg)(0.44×103J/kgK))[(-20.0°C)10.0+273.15K+(20.0°C)70.0+273.15K]=2.7J/K

The change in entropy of universe is 2.7J/K.

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Chapter 15 Solutions

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