PART 1: An object at 20?C absorbs 25.0 J of heat. What is the change in entropy
ID: 2284084 • Letter: P
Question
PART 1: An object at 20?C absorbs 25.0 J of heat. What is the change in entropy ?S of the object? PART 2: An object at 500 K dissipates 25.0 kJ of heat into the surroundings. What is the change in entropy ?S 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 3: An object at 400 K absorbs 25.0 kJ of heat from the surroundings. What is the change in entropy ?S 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 4: 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 ?Ssys 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.
Explanation / Answer
entropy change = change in heat content /constant temp (K)
(?S) = ?Q/T = Joule/kelvin
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A) absorbs heat > ?Q = + 25 J
T = 20+273 = 293 K
?S = ?Q/T = 25/293 = +0.85 J/K >> increase in entopy of object
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B) loses heat > ?Q = - 25000 J
constant T = 500 K
?S = ?Q/T = 25000/500 = - 50 J/K >> decrease in entopy of object
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C) closed system >> isolated from rest of unioverse
object 1 > absorbs heat > ?Q1 = + 25000 J
T1 = 400 K
?S1 = ?Q1/T1 = + 25000/400 J/K >> increase in entopy of object1
object 2 > loses same heat > ?Q2 = - 25000 J
T2 = 500 K
?S2 = ?Q2/T2 = - 25000/500 J/K >> increase in entopy of object2
net change in entropy of system >
?S = ?S1 + ?S2 = 25000/400 - 25000/500
?S = 250/4 - 250/5 = 250[0.25 - 0.20] = +12.5 J/K
net entropy of system increases by 12.5 J/K
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