Calculation of entropy change

In summary, the heat capacity of the cube is equal to the heat capacity of the reservoir, and the entropy change of the universe is zero after the heat transfer is complete.
  • #1

Homework Statement


A solid metallic cube of heat capacity S is at temperature 300K. It is brought in contact with a reservoir at 600K. If the heat transfer takes place only between the reservoir and the cube, the entropy change of the universe after reaching the thermal equillibrium is

A. 0.69S
B. 0.54S
C. 0.27S
D. 0.19S

[Answer : 0.19S]

Homework Equations


(heat supplied)=SΔT

Q = SΔT

Change in entropy = (change in heat Q)/T

ΔE = Q/T

[I have taken entropy as E rather than usual S since S is already taken for heat capacity]

The Attempt at a Solution


[/B]
Should I take reservoir as an infinite pool of temperature? Then, I get

Q = SΔT
Q=S*(600-300) since, at thermal equillibrium the temperatures are same
Q=300S
ΔE=(300/600)=0.5 which is not the answer

If I take reservoir which loses temperature, I am unable to continue with the problem since I do not know the heat capacity of it.

Is there a different approach to this problem?
 
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  • #2
astrophysics12 said:

Homework Statement


A solid metallic cube of heat capacity S is at temperature 300K. It is brought in contact with a reservoir at 600K. If the heat transfer takes place only between the reservoir and the cube, the entropy change of the universe after reaching the thermal equillibrium is

A. 0.69S
B. 0.54S
C. 0.27S
D. 0.19S

[Answer : 0.19S]

Homework Equations


(heat supplied)=SΔT

Q = SΔT

Change in entropy = (change in heat Q)/T

ΔE = Q/T

[I have taken entropy as E rather than usual S since S is already taken for heat capacity]

The Attempt at a Solution


[/B]
Should I take reservoir as an infinite pool of temperature? Then, I get

Q = SΔT
Q=S*(600-300) since, at thermal equillibrium the temperatures are same
Q=300S
ΔE=(300/600)=0.5 which is not the answer

If I take reservoir which loses temperature, I am unable to continue with the problem since I do not know the heat capacity of it.

Is there a different approach to this problem?

Bits are right, but a lot of what you have is incorrect.

BTW this is a HORRIBLE problem -- it seems like the person who wrote the problem is trying to confuse you with the heat capacity = "S' garbage.

dS = dqrev/T

For the block, to calculate Delta S, you will need to integrate the above over the temperature range.

For the bath, assuming that the bath has a huge heat capacity, and does not change temperature (T is constant), then Delta S bath = Qbath/T

To find Qbath, you need to find Qblock (Qbath = - Qblock -- energy is conserved). Qblock can be found from "S" Delta T, as you have noted.

Delta S Universe = Delta S block + Delta S bath
 
  • #3
Thanks a lot. I got it.

This is actually from a question paper that I was practicing.
 

1. What is entropy change?

Entropy change refers to the measure of the disorder or randomness in a system. It is a thermodynamic property that describes the level of energy dispersal or the number of ways in which a system can be arranged.

2. How is entropy change calculated?

Entropy change can be calculated using the equation ΔS = Qrev/T, where ΔS is the change in entropy, Qrev is the reversible heat transfer, and T is the temperature in Kelvin.

3. What is the relationship between entropy change and energy?

Entropy change is directly related to energy. As the energy in a system increases, the disorder and randomness also increase, resulting in a higher entropy change.

4. What factors affect entropy change?

Entropy change is affected by temperature, pressure, and the number of particles in a system. Generally, an increase in temperature and pressure leads to an increase in entropy change, while a decrease in the number of particles leads to a decrease in entropy change.

5. Why is entropy change important in thermodynamics?

Entropy change is a crucial concept in thermodynamics because it helps us understand how energy is distributed and transformed within a system. It also plays a significant role in determining the direction and spontaneity of chemical reactions and physical processes.

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