Thermodynamics and Entropy- reservoir and block problem

In summary, a 382 g block at a lower temperature than a thermal reservoir experiences reversible heat transfer until thermal equilibrium is reached. The specific heat of the block can be calculated using the change in entropy and temperatures provided in Figure 20-22. However, it is important to note that the initial temperature is not equivalent to Ta as shown on the graph.
  • #1
emmy
37
0

Homework Statement


A 382 g block is put in contact with a thermal reservoir. The block is initially at a lower temperature than the reservoir. Assume that the consequent transfer of energy as heat from the reservoir to the block is reversible. Figure 20-22 gives the change in entropy ΔS of the block until thermal equilibrium is reached. The scale of the horizontal axis is set by Ta = 280 K and Tb = 520 K. What is the specific heat of the block?



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Homework Equations


ΔS= mCspln(T/To)

The Attempt at a Solution


ΔS= mCspln(T/To)

Csp=ΔS/(m*ln(T/To))

where
ΔS= 100 J/K
m= .382 kg
T=520K
To=280K

plug and chug,
Csp= 100/(.382*ln(520/280)) =422.88J/K*kg

where did I go wrong ):
 
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  • #2
your T0 is not Ta

look carefully at the graph^^
 
  • #3
quietrain said:
your T0 is not Ta

look carefully at the graph^^


Ohhhhh my gosh this is why i miss points on exams lol! Thanks c:
 

FAQ: Thermodynamics and Entropy- reservoir and block problem

1. What is the difference between a reservoir and a block in thermodynamics?

A reservoir in thermodynamics refers to a large system that is able to exchange energy and matter with its surroundings. It is considered to have an infinite capacity and does not change its state during a thermodynamic process. On the other hand, a block is a smaller system that is in contact with the reservoir and exchanges energy and matter with it. The block is considered to be a closed system and can change its state during a thermodynamic process.

2. How does the concept of entropy apply to a reservoir and block problem?

Entropy is a measure of the disorder or randomness of a system. In a reservoir and block problem, the entropy of the overall system will tend to increase over time due to the exchange of energy and matter between the reservoir and the block. This is because the block will tend to become more disordered as it absorbs energy from the reservoir, which will decrease in temperature and become more ordered.

3. Can a block be considered a reservoir in a thermodynamics problem?

No, a block cannot be considered a reservoir in a thermodynamics problem. A reservoir is typically a much larger system with an infinite capacity, whereas a block is a smaller system that can change its state during a thermodynamic process. Additionally, a reservoir is in thermal equilibrium, meaning its temperature remains constant, while a block can change its temperature during a process.

4. What is the significance of a reversible process in a reservoir and block problem?

In a reservoir and block problem, a reversible process is one in which the system can return to its original state without leaving any net effect on the surroundings. This is important because it allows for the maximum possible efficiency in the exchange of energy and matter between the reservoir and the block. In contrast, an irreversible process will result in a decrease in the overall entropy of the system.

5. How does the first and second law of thermodynamics apply to a reservoir and block problem?

The first law of thermodynamics, also known as the law of conservation of energy, states that energy cannot be created or destroyed, only transferred or converted. In a reservoir and block problem, this law applies as energy is exchanged between the two systems. The second law of thermodynamics states that the total entropy of an isolated system will always tend to increase over time. In a reservoir and block problem, this law applies as the overall entropy of the system will tend to increase due to the exchange of energy and matter between the reservoir and the block.

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