Question

A hot object and a cold object are placed in thermal contact and the combination is isolated. They transfer energy until they reach a common temperature. The change ΔSh in the entropy of the hot object, the change ΔSc in the entropy of the cold object, and the change ΔStotal in the entropy of the combination are:
a) ΔSh > 0, ΔSc > 0, ΔStotal > 0 b) ΔSh < 0, ΔSc > 0, ΔStotal > 0 c) ΔSh < 0, ΔSc > 0, ΔStotal < 0 d) ΔSh > 0, ΔSc < 0, ΔStotal > 0 e) ΔSh > 0, ΔSc < 0, ΔStotal < 0

Answer 1

Answer: b) $\Delta Sh$, $\Delta Sc>0$, $\Delta Stotal>0$

Explanation:

According to the second principle of thermodynamics:  

"The amount of entropy in the universe tends to increase over time ".

Entropy is a thermodynamic quantity defined as a criterion to predict the evolution or transformation of thermodynamic systems, it is also the measure of the disorder of a system and is a function of state. Therefore it is used to measure the degree of organization of a system.  

In addition, entropy depends on the temperature, if temperature increases, entropy increases as well; if temperature decreases, entropy decreases as well, in fact, if a substance reaches absolute zero, the entropy reaches a minimum and constant value. This means every chemical and thermal reaction must increase the total entropy of the universe.  

So, in the case of the hot and the cold object placed in thermal contact and the combination is isolated:

-As the hot object decreases its temperature in the thermal transfer, its entropy also decreases ($\Delta Sh$).

-As the cold object gains temperature in the thermal transfer, its entropy also increases ($\Delta Sc>0$).

Finally, remembering that any thermal reaction must increase the total entropy of the universe, the resultant total entropy of the system will also be increased ($\Delta Stotal>0$)