 # Entropy homework Anonymous

### Question Description

1. In question 1, we consider processes similar to Process A from the tutorial, with a block with heat capacity 400 J/K and initial temperature 280 K and a second block with heat capacity 200 J/K and initial temperature 340 K.

A. In process B, the 280 K block and 340 K block change in temperature, but in the opposite direction as in process A. In other works, the temperature of the 280 K block decreases to 260 K and the 340 K block increases to 380 K. The blocks are still thermally isolated from all other objects.

1. Is process B allowed by the first law of thermodynamics? Explain your reasoning.

2. What is the total change in entropy of the two-block system during process B?

B. In process C, both blocks begin at 300 K but their temperatures diverge so that the block with heat

capacity 400 J/K ends up at a temperature of 280 K and the other ends up at 340 K. (Note that this

process is the reverse of process A).

1. Is process C allowed by the first law of thermodynamics? Explain your reasoning.

2. Using your result from part A.5, find the change in entropy for each block and for the two block system during process C.

C. The following questions refer to processes A (from the tutorial), B, and C.

1. For those processes that you would expect to observe, was the total change in entropy of all systems involved greater than, less than, or equal to zero?

2. For those processes that you would not expect to observe, was the total change in entropy of all systems involved greater than, less than, or equal to zero?

2. Consider a macroscopic system of two blocks that are in thermal contact. The two-block system is isolated from the rest of the universe. Initially, a hotter block is at a temperature TH and a colder block is at a temperature TC. In the final state of the system, the two blocks are at the same intermediate temperature TI.

A. Compare the following quantities for the initial state and final state: Entropy for the block that starts at TH

Sf > Si Sf < Si Si = Sf not enough information

Entropy for the block that starts at TL

Sf > Si Sf < Si Si = Sf not enough information

Combined entropy for the two block system

Sf > Si Sf < Si Si = Sf not enough information

B. Consider the following statement:

“When the blocks approach equilibrium, they move to a final state that is more natural, so their level

of order increases.”

Do you agree with this statement?

Does this statement agree with the second law of thermodynamics?

3. An ice cube with a mass of 30.0 g is dropped into a beaker of water. The initial temperature of the ice cube is 0° C and the initial temperature of the water in the beaker is 20° C. The heat capacity of the water in the beaker is 2000 J/K. The latent heat of fusion for water is 333 J/g. Assume that the system consisting of the beaker and the cube does not thermally interact with its surroundings.

A. Determine the equilibrium temperature of the system.

B. Find the changes in entropy for the following processes and carefully specify the system (or subsystem) for which you found ΔS.

a. the melting of the ice cube: (Hint: Does the temperature of the ice change during this process?)

b. the cooling of the water as the ice cube is melting:

c. the heating of the melted ice to its final temperature:

d. the cooling of the warm water to its final temperature:

C. Find the total change in entropy during the process:

D. Would you expect the reverse of this process (i.e., 30 g of ice forms spontaneously from a beaker of water at the final temperature) to occur? Is the reverse of this process forbidden by the first or second law of thermodynamics?

HW file is also being attached below along with the tutorial paper too.

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Ace_Tutor
School: UC Berkeley  solution attached

QUESTION 1:
A)
1. The process B is allowed by the first law of thermodynamics because

work done by block A  work done by block B
  400 J / K  280 K  260 K    200 J / K  380 K  340 K 

 8000 J  8000 J
2. The total change in entropy of the two-block system during process is calculated by

T 
T 
S  C A ln  A1   CB ln  B1 
 TA2 
 TB 2 
 260 
 380 
 400 ln 
  200 ln 

 280 
 340 
 7.40
B)
1. Process C is not allowed by the first law of thermodynamics since

work done by block C =
=  400 J / K  340 K  280 K 
 24000 J
which is not the same work done of 8000 J as in the answer in part A.
2. The total change in entropy of each block is calculated by
For block 1:

T 
 280 K 
S1  C ln  1 f    400 J / K  ln 
  27.60
 300K 
 T1i 
For block 2:

T 
 340 K 
S1  C ln  1 f    400 J / K  ln 
  50.07
 300 K 
 T1i 
The two-block system during process is calculated by

S  S1  S2  27.60  50.07  22.47

C)
1. The total change in entropy of all systems involved was less than zero as ...

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