A metal rod is used to conduct heat between two reservoirs at temperatures Th and Tc, respectively. When an amount of heat Q flows through the rod from the hot to the cold reservoir, what is the net entropy change of the rod, the hot reservoir, the cold reservoir, and the universe? (Use any variable or symbol stated above as necessary.)

ΔS(rod) = 0 J/K    

ΔS(hot) = -Q/T_h

ΔS(cold) =  +Q/T_c  

ΔS(universe) = -Q/T_h  + Q/T_c  

Explanation:

We know that an isolated system tends toward disorder and entropy is the measure of this disorder. This means that the entropy of the universe increases in all real processes. And we know that the change in entropy for a system and its surroundings is always positive for an irreversible process like our situation here (in this example). So that, ΔS > 0  

ΔS(universe) = ΔS(hot) +ΔS(cold) + ΔS(rod)

Now we need to find the entropy change of the rod. The rod already connected and already in a dynamically steady-state condition and there is no state change occurs in the rod in this thermodynamic processes. Note that the rod absorbs some heat Q from the hot reservoir and at the same time exhausts the same amount of heat to the cold reservoir. This means that the rod is in a dynamic thermal equilibrium and that its temperature will remain constant.  

Therefore,  

ΔS(rod) = +Q/T(rod) ÷ -Q/T(rod)

ΔS(rod) = 0 J/K                                 (Metal Rod)  

Because the temperature of the reservoirs are constant (and we are supposing that the process of losing or adding heat to the metal rod could be reversible in another theoretical experiment), we will use the formula of ΔS = Q/T

Therefore,  

ΔS(hot) = -Q/T_h                                 (Hot Reservoir)  

Because the hot reservoir is losing heat, we put the negative sign.  

ΔS(cold) =  +Q/T_c                             (Cold Reservoir)  

Because the cold reservoir is absorbing heat, we put the positive sign.  

Because T_h   > T_c  , so that, I ΔS(hot) I < ΔS(cold) . This means that the decrease of the entropy of the hot reservoir is less than the increase of the entropy of the cold reservoir.  

Therefore, the net entropy change of the universe is greater than zero.  

ΔS(universe) = ΔS(hot) +ΔS(cold) + ΔS(rod)

ΔS(universe) = ΔS(hot) +ΔS(cold) + 0

ΔS(universe) = -Q/T_h  + Q/T_c       (which is greater than zero)  

1 the 4 is sig

when two balls of the different masses are dropped from equal height, they reach the ground at the same time .this is because they experience the same acceleration.

the acceleration is independent of mass.

thus when they are dropped they will always maintain the same velocity and travel the same distance (neglect air resistance of course) at the same time.