Required heat, entropy change of object dropped in water

AI Thread Summary
To determine the heat needed to raise the temperature of argon from 303K to 323K, the specific heat capacity at constant volume (c_v) is used, calculated as c_v=3/2*R. The entropy change for the reservoir is negative, indicating a reduction in entropy, while the system's entropy change is positive, resulting in a total universe entropy change of 0.02. The number of moles (n) of argon is calculated to be 0.04 using the ideal gas law, which is essential for accurate entropy calculations. It is confirmed that n should be included in the entropy equations for both the reservoir and the system. The discussion emphasizes the importance of correctly applying thermodynamic principles to solve the problem.
ElectricEel1
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Homework Statement


A 1 liter container is filled with argon to pressure of 10^5Pa at 303K. Dropped into pool at temp 323K. How much heat is needed to heat the gas to 323K? what is the entropy change in the gas and the universe? ignore the entropy change in container.

Homework Equations


c_v=3/2*R
dq=cdt
deltaS=integral of dQ/T

The Attempt at a Solution


I started with the entropy:

reservoir:
T is constant so delta Q/ T_R = C(T_s-T_r)/T_r = 0.77. negative sign because its entropy reduced.
system:
evaluating the same integral with a non constant T gave me C*ln(T_r/T_s) = 0.79.
Universe:
change in the universes entropy was 0.02 after adding both of these together.

I'm unsure of these answers and at first I figured I should just use C(T_s-T_r) to find the heat added but the question mentions the initial pressure and the constant volume of 1 litre so I think I am missing something.
Thanks
 
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I figured I needed to find n so I calculated n to be 0.04 from pV=nRT.
Does this mean I should have had n before C in the entropy parts too?
 
ElectricEel1 said:
I figured I needed to find n so I calculated n to be 0.04 from pV=nRT.
Does this mean I should have had n before C in the entropy parts too?
Yes. You've done great.
 
Chestermiller said:
Yes. You've done great.
thank you sir
 
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