Answer:
0.00417 kW/K or 4.17 W/K
Second law is satisfied.
Explanation:
Parameters given:
Rate of heat transfer, Q = 2kW
Temperature of hot reservoir, Th = 800K
Temperature of cold reservoir, Tc = 300K
The rate of entropy change is given as:
ΔS = Q * [(1/Tc) - (1/Th)]
ΔS = 2 * (1/300 - 1/800)
ΔS = 2 * 0.002085
ΔS = 0.00417 kW/K or 4.17 W/K
Since ΔS is greater than 0, te the second law of thermodynamics is satisfied.
The entropy change is greater than zero and it satisfies the second law of thermodynamics.
Given data:
The heat transfer rate from the reservoir is, Q = 2 kW = 2000 W.
The temperature of hot reservoir is, T = 800 K.
The temperature of cold reservoir is, T' = 300 K.
As per the second law of thermodynamics, when the entropy of system is greater than the zero, s > 0, then the system will follow the second law of thermodynamics.
Then the expression for the rate of entropy change is given as,
ΔS = Q [(1/T') - (1/T)]
ΔS = 2 (1/300 - 1/800)
ΔS = 2 (0.002085)
ΔS = 0.00417 kW/K or 4.17 W/K
Clearly, ΔS is greater than 0, then the second law of thermodynamics is satisfied.
Thus, we can conclude that the entropy change is greater than zero and it satisfies the second law of thermodynamics.
Learn more about the second law of thermodynamics here:
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