Question

Two hundred joules of heat are removed from a heat reservoir at a temperature of 200 K. What is the entropy change of the reservoir?

a) 0.002 kJ/K
b) 0.001 kJ/K
c) 0.0025 kJ/K
d) 0.0015 kJ/K

Answer 1

The entropy change of the heat reservoir is -0.001 kJ/K when 200 joules of heat are removed at a temperature of 200 K. The negative sign indicates that entropy is being removed from the system.

Step-by-step explanation:

The question deals with the concept of entropy change in a thermodynamic process. Specifically, it asks for the entropy change of a heat reservoir from which 200 joules of heat are removed at a constant temperature of 200 K. The formula for calculating entropy change (ΔS) in a reversible process is ΔS = Q/T, where Q is the heat exchanged and T is the temperature in kelvins.

To answer the question, we calculate the entropy change as follows:

ΔS = -Q/T

ΔS = -200 J / 200 K

ΔS = -1 J/K

ΔS = -0.001 kJ/K

Therefore, the entropy change of the reservoir is -0.001 kJ/K, which corresponds to option b).