Quick Physics Problem...?

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

Hi, i hope this would help.

Firstly, entropy is a measure of orderliness of a system and the greater the physical and thermal chaos, the more disorderly the system is , the higher the entropy value. Entropy is also a temperature dependant quantity.

when the two objects mentioned are in contact, the hot object will lose heat while the cold object will gain heat until thermal equilibrium is reached.

For the hot object, the temperature decreases as it loses heat. When temperature is decreased, disorder generally decreases because physical chaos and temperature chaos both decreases. thus entropy decreases.(i.e delta SH <0)

For the cold object, it will gain heat from the hot object. As a result disorder will increase as physical chaos and thermal chaos both increase. Thus the entropy increases. (i.e. delta SC >0)

the total change in entropy will be >0 as physical chaos still exist because the particles in the objects are still vibrating.

Therefore, i think that the most suitable answer to this question is b.

Hope it helps!

Answer 2

The hot object loses entropy. negative
The cold object gains entropy. positive
The net exchange (Stotal) must be positive.

Answer b. I think.

Double checked my thermodynamics book for this equation.
In an isolated system
Sgen = deltaStotal = deltaSsystem + deltaSsurroundings > or = 0 .
Found this website. Hope this helps.
http://www.entropysite.com/students_approach.html

Answer 3

b.

Doug