An object at 20∘C absorbs 25.0 J of heat. What is the change in entropy ΔS of the object? Express your answer numerically in joules per kelvin. ΔS = nothing J/K Request Answer Part E An object at 500 K dissipates 25.0 kJ of heat into the surroundings. What is the change in entropy ΔS of the object? Assume that the temperature of the object does not change appreciably in the process. Express your answer numerically in joules per kelvin. ΔS = nothing J/K Request Answer Part F An object at 400 K absorbs 25.0 kJ of heat from the surroundings. What is the change in entropy ΔS of the object? Assume that the temperature of the object does not change appreciably in the process. Express your answer numerically in joules per kelvin. ΔS = nothing J/K Request Answer Part G Two objects form a closed system. One object, which is at 400 K, absorbs 25.0 kJ of heat from the other object, which is at 500 K. What is the net change in entropy ΔSsys of the system? Assume that the temperatures of the objects do not change appreciably in the process. Express your answer numerically in joules per kelvin.

Question:

Part D) An object at 20∘C absorbs 25.0 J of heat. What is the change in entropy ΔS of the object? Express your answer numerically in joules per kelvin.

Part E) An object at 500 K dissipates 25.0 kJ of heat into the surroundings. What is the change in entropy ΔS of the object? Assume that the temperature of the object does not change appreciably in the process. Express your answer numerically in joules per kelvin.

Part F) An object at 400 K absorbs 25.0 kJ of heat from the surroundings. What is the change in entropy ΔS of the object? Assume that the temperature of the object does not change appreciably in the process. Express your answer numerically in joules per kelvin.

Part G) Two objects form a closed system. One object, which is at 400 K, absorbs 25.0 kJ of heat from the other object,which is at 500 K. What is the net change in entropy ΔSsys of the system? Assume that the temperatures of the objects do not change appreciably in the process. Express your answer numerically in joules per kelvin.

D) 85 J/K

E) - 50 J/K

F) 62.5 J/K

G) 12.5 J/K

Explanation:

Let's make use of the entropy equation: ΔS =

Part D)

Given:

T = 20°C = 20 +273 = 293K

Q = 25.0 kJ

Entropy change will be:

ΔS =

= 85 J/K

Part E)

Given:

T = 500K

Q = -25.0 kJ

Entropy change will be:

ΔS =

= - 50 J/K

Part F)

Given:

T = 400K

Q = 25.0 kJ

Entropy change will be:

ΔS =

= 62.5 J/K

Part G:

Given:

T1 = 400K

T2 = 500K

Q = 25.0 kJ

The net entropy change will be:

ΔS =

= 12.5 J/K

to be fair, you have to have a very high iq to understand rick and morty. the humour is extremely subtle, and without a solid grasp of theoretical physics most of the jokes will go over a typical viewer's head. there's also rick's nihilistic outlook, which is deftly woven into his characterisation- his personal philosophy draws heavily from narodnaya volya literature, for instance. the fans understand this stuff; they have the intellectual capacity to truly appreciate the depths of these jokes, to realise that they're not just funny- they say something deep about life. as a consequence people who dislike rick & morty truly are idiots- of course they wouldn't appreciate, for instance, the humour in rick's existential catchphrase "wubba lubba dub dub," which itself is a cryptic reference to turgenev's russian epic fathers and sons. i'm smirking right now just imagining one of those addlepated simpletons scratching their heads in confusion as dan harmon's genius wit unfolds itself on their television screens. what fools.. how i pity them.

and yes, by the way, i do have a rick & morty tattoo. and no, you cannot see it. it's for the ladies' eyes only- and even then they have to demonstrate that they're within 5 iq points of my own (preferably lower) beforehand. nothin personnel kid