Required heat, entropy change of object dropped in water

In summary, the task is to determine the amount of heat needed to heat argon gas in a 1 liter container from 303K to 323K, as well as the entropy change in the gas and the universe. The equation c_v=3/2*R is used to find the heat added, and the equation deltaS=integral of dQ/T is used for entropy calculations. The correct value for entropy change in the universe is 0.02, which is found by adding the entropy changes for the system and the reservoir. The value of n is needed in the entropy calculations and is calculated to be 0.04 from the ideal gas law.
  • #1
ElectricEel1
51
0

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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  • #2
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?
 
  • #3
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.
 
  • #4
Chestermiller said:
Yes. You've done great.
thank you sir
 

1. What is required heat?

Required heat is the amount of energy needed to raise the temperature of an object by a certain amount. It is typically measured in joules or calories.

2. How does an object being dropped in water affect its entropy?

When an object is dropped in water, its entropy will increase. This is because the object's molecules will become more dispersed and disordered as they mix with the water molecules.

3. What factors affect the entropy change of an object dropped in water?

The entropy change of an object dropped in water is affected by the temperature difference between the object and the water, the mass and specific heat of the object, and the temperature and volume of the water.

4. How is the required heat and entropy change of an object dropped in water related?

The required heat and entropy change of an object dropped in water are related through the equation Q = mCΔT, where Q is the required heat, m is the mass of the object, C is its specific heat, and ΔT is the change in temperature. This equation shows that as the required heat increases, so does the entropy change.

5. How can the required heat and entropy change of an object dropped in water be calculated?

The required heat and entropy change of an object dropped in water can be calculated using the equation Q = mCΔT and the formula ΔS = Q/T, where Q is the required heat, m is the mass of the object, C is its specific heat, ΔT is the change in temperature, and T is the temperature of the water. Alternatively, these values can also be calculated using experimental data and the principles of thermodynamics.

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