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An amount of heat 1000kJ is transferred from a heat reservoir at temperature TA to another heat reservoir at temperature TB . If TA = 800K and and TB = 500K, what is the change in entropy ΔS of the combined system in JK-1
Correct answer is '750'. Can you explain this answer?
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An amount of heat 1000kJ is transferred from a heat reservoir at tempe...
Conventionally, heat given to the system is taken to be positive and heat extracted from the system is taken to be negative,
ΔS = ΔSA + ΔS
ΔS = 750JK-1
The correct answer is: 750
ΔS = ΔSA + ΔS
B
ΔS = 750JK-1
The correct answer is: 750
Most Upvoted Answer
An amount of heat 1000kJ is transferred from a heat reservoir at tempe...
Calculation of Change in Entropy (ΔS) of the Combined System:
1. Calculate the Change in Entropy:
- The change in entropy (ΔS) can be calculated using the formula: ΔS = Q / T, where Q is the heat transferred and T is the temperature.
2. Calculate the Heat Transferred:
- The heat transferred (Q) is given as 1000 kJ.
3. Calculate the Temperatures in Kelvin:
- The initial temperature (TA) is 800K and the final temperature (TB) is 500K.
4. Calculate the Change in Entropy:
- Using the formula, ΔS = Q / T, we get:
ΔS = 1000 kJ / 800K - 1000 kJ / 500K
ΔS = 1.25 kJ/K - 2 kJ/K
ΔS = -0.75 kJ/K
- Since entropy is a state function, the change in entropy is the sum of the changes of the two reservoirs, so the absolute value of -0.75 kJ/K is taken.
ΔS = 0.75 kJ/K = 750 J/K
Therefore, the change in entropy (ΔS) of the combined system is 750 J/K.
1. Calculate the Change in Entropy:
- The change in entropy (ΔS) can be calculated using the formula: ΔS = Q / T, where Q is the heat transferred and T is the temperature.
2. Calculate the Heat Transferred:
- The heat transferred (Q) is given as 1000 kJ.
3. Calculate the Temperatures in Kelvin:
- The initial temperature (TA) is 800K and the final temperature (TB) is 500K.
4. Calculate the Change in Entropy:
- Using the formula, ΔS = Q / T, we get:
ΔS = 1000 kJ / 800K - 1000 kJ / 500K
ΔS = 1.25 kJ/K - 2 kJ/K
ΔS = -0.75 kJ/K
- Since entropy is a state function, the change in entropy is the sum of the changes of the two reservoirs, so the absolute value of -0.75 kJ/K is taken.
ΔS = 0.75 kJ/K = 750 J/K
Therefore, the change in entropy (ΔS) of the combined system is 750 J/K.
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An amount of heat 1000kJ is transferred from a heat reservoir at temperature TA to another heat reservoir at temperature TB . If TA = 800K and and TB = 500K, what is the change in entropy ΔS of the combined system in JK-1Correct answer is '750'. Can you explain this answer?
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