How to Calculate Entropy Change for Different Scenarios Involving a Can of Coke?

[Discostu19]

Heat and Entropy Change HELPPPP!

Calculate change in entropy of the universe for:
1] A can of coke, at initial temperature of 5 degrees C, with heat capacity 2500JK-1, is allowed to reach thermal equilibrium with its surroundings at 40 degrees C

2] Two Cans of coke at 5 degrees C and 40 degrees C are placed in thermal contact with each other inside a cool bag where they are thermally insulated from their surroundings and allowed to equilibrate

3]One of the above coke cans at 40 degrees C weighing 250g is dropped from a height of 50m into a shallow pond of negligible depth which is also at 40 degrees C


Attempt

1] using delta S = C_pln(T_f/T_i)
giving 2500xln(40/5)
=5198.6JK^-1

ii]
using T_f=(C₁T₁+C₂T₂)/C₁C₂

[2500x40]+[2500x5]/2500+2500
=22.5 degrees C

therefore delta S = C₁ln(T_f/T₁)+C₂ln(T_f/T₂)


=2500ln(22.5/40)+2500ln(22.5/5)
=2321.8

so far so gd?

for iii] do i use delta S = P.E / temp of surrounding ?

 

[MathematicalPhysicist]

For 1), you need to change the tempratures to kelvin units.

For 3), if I understand the question correctly, you use the connection between energy and heat transfer, Q=U+W, U=mgh, the work that being done by the bodies, W, not sure how to get it.
Anyway, dQ=SdT+TdS, there isn't a change in the tempratures, and the change is entropy is what you're after, so dQ=TdS, dS=dQ/T, find the change in the heat transfer and I believe your'e done, not sure though.

 

[Discostu19]

cheers :)