Question:
Part D) An object at 20∘C absorbs 25.0 J of heat. What is the change in entropy ΔS of the object? Express your answer numerically in joules per kelvin.
Part E) An object at 500 K dissipates 25.0 kJ of heat into the surroundings. What is the change in entropy ΔS of the object? Assume that the temperature of the object does not change appreciably in the process. Express your answer numerically in joules per kelvin.
Part F) An object at 400 K absorbs 25.0 kJ of heat from the surroundings. What is the change in entropy ΔS of the object? Assume that the temperature of the object does not change appreciably in the process. Express your answer numerically in joules per kelvin.
Part G) Two objects form a closed system. One object, which is at 400 K, absorbs 25.0 kJ of heat from the other object,which is at 500 K. What is the net change in entropy ΔSsys of the system? Assume that the temperatures of the objects do not change appreciably in the process. Express your answer numerically in joules per kelvin.
Answer:
D) 85 J/K
E) - 50 J/K
F) 62.5 J/K
G) 12.5 J/K
Explanation:
Let's make use of the entropy equation: ΔS = [tex] \frac{Q}{T} [/tex]
Part D)
Given:
T = 20°C = 20 +273 = 293K
Q = 25.0 kJ
Entropy change will be:
ΔS = [tex] \frac{25*1000}{293} [/tex]
= 85 J/K
Part E)
Given:
T = 500K
Q = -25.0 kJ
Entropy change will be:
ΔS = [tex] \frac{-25*1000}{500} [/tex]
= - 50 J/K
Part F)
Given:
T = 400K
Q = 25.0 kJ
Entropy change will be:
ΔS = [tex] \frac{25*1000}{400} [/tex]
= 62.5 J/K
Part G:
Given:
T1 = 400K
T2 = 500K
Q = 25.0 kJ
The net entropy change will be:
ΔS = [tex] (\frac{25*1000}{400}) + (\frac{-25*1000}{500}[/tex]
= 12.5 J/K
Answer:
A) The change in entropy [tex]\delta S = 0.085J/K[/tex]
B) The change in entropy [tex]\delta S = -50J/K[/tex]
C) The change in entropy [tex]\delta S = 62.5J/K[/tex]
D) The net change in entropy [tex]\delta S = 12.5J/K[/tex]
Explanation:
A)
[tex]T = 20^oC + 273k\\\\T = 293k[/tex]
expression for change in entropy,
[tex]\delta S = \frac{\delta Q}{T}\\\\\delta S = \frac{25}{293}\\\\\delta S = 0.085J/K[/tex]
B) [tex]\delta S = \frac{\delta Q}{T}\\\\\delta S = \frac{-25*10^3}{500}\\\\\delta S = -50J/K[/tex]
The negative sign indicates the heat lost into the surrounding
C)
[tex]\delta S = \frac{\delta Q}{T}\\\\\delta S = \frac{25*10^3}{400}\\\\\delta S = 62.5J/K[/tex]
The entropy remains constant in the adiabatic process because no heat is given to the system in this process.
D)
[tex]\delta S_1 = \frac{\delta Q_1}{T_1}\\\\\delta S_1 = \frac{25*10^3}{400}\\\\\delta S_1 = 62.5J/K[/tex]
similarly,
[tex]\delta S_2 = \frac{\delta Q_2}{T_2}\\\\\delta S_2 = \frac{25*10^3}{500}\\\\\delta S_2 = 50J/K[/tex]
Therefore the net change in entropy is,
[tex]\delta S = \delta S_1 - \delta S_2\\\\\delta S = 62.5 - 50\\\\\delta S = 12.5J/K[/tex]
Entropy is not a conserved quantity because it can be created but cannot be destroyed.
The net change in entropy is calculated by difference of the change in entropy at two different temperatures.
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