Cherry Pie Entropy: Find Final Temp from Heat & Entropy Increase

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In summary, the problem involves determining the final temperature of a cherry pie that initially has 1.70 ✕ 105 J of heat entering it at 20.0°C, resulting in an entropy increase of 470 J/K. Using the equations S = Q/T and Q = C(T2-T1), the final temperature can be found by setting these equations equal to each other and solving for T2. The heat capacity, C, cancels out, leaving the equation T2 = T1 + Q/S. By plugging in the values from the problem statement, the final temperature is determined to be 293.4 K.
  • #1
suchjoe
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Homework Statement



When 1.70 ✕ 105 J of heat enters a cherry pie initially at 20.0°C, its entropy increases by 470 J/K. What is its final temperature?

Homework Equations





The Attempt at a Solution


I have no clue
 
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  • #2
suchjoe said:

Homework Statement



When 1.70 ✕ 105 J of heat enters a cherry pie initially at 20.0°C, its entropy increases by 470 J/K. What is its final temperature?

Homework Equations





The Attempt at a Solution


I have no clue
Sorry, but you have to try something. Do you know the equation for how the heat that enters is related to the temperature rise? Do you know the equation for how the entropy change is related to the initial and final temperatures. Both these equations involve the heat capacity.
 
  • #3
suchjoe said:

Homework Statement



When 1.70 ✕ 105 J of heat enters a cherry pie initially at 20.0°C, its entropy increases by 470 J/K. What is its final temperature?

Homework Equations


The Attempt at a Solution


I have no clue

Generally, S = Q/T. Kind of simplistic, but that is probably what you're going to use to find the net heat transfer into the cherry pie. Then, you'll need to know the heat capacity of the cherry pie to determine the change in temperature. Does the problem statement give the temperature of the surroundings? If so, I'd use that as the temp at which the heat transfer occurs. And if this is a college thermo class, dS = Cavg*ln(t2/t1)
 
  • #4
wreckemtech said:
Generally, S = Q/T. Kind of simplistic, but that is probably what you're going to use to find the net heat transfer into the cherry pie. Then, you'll need to know the heat capacity of the cherry pie to determine the change in temperature. Does the problem statement give the temperature of the surroundings? If so, I'd use that as the temp at which the heat transfer occurs. And if this is a college thermo class, dS = Cavg*ln(t2/t1)
Something like the last equation is the one to use, but I'm sure they want you to assume that the heat capacity is constant. You don't need to know the surroundings temperature, so don't waste your time looking for that.

So again, in terms of the mass of the pie, the heat capacity, and the initial and final temperatures, what is the change in entropy? You don't need to know the mass or the heat capacity, because they are going to cancel out. And again, in terms of the mass of the pie, the heat capacity, and the initial and final temperatures, what is the amount of heat that enters? Can you combine these two equations?

Chet
 
  • #5
You know, I somehow completely missed the fact that Q was given in the problem statement. I thought the poster was only given T1 and dS. If Q is known, it should be easy enough to substitute something like Q=C(T2-T1) into the first equation, and then C will cancel.
 
  • #6
wreckemtech said:
You know, I somehow completely missed the fact that Q was given in the problem statement. I thought the poster was only given T1 and dS. If Q is known, it should be easy enough to substitute something like Q=C(T2-T1) into the first equation, and then C will cancel.
According to PF rules, we really shouldn't be doing the problem for suchjoe. We should just be giving him little hints so that he can analyze it for himself. I feel like I've already revealed too much about how to do it.
 
  • #7
Sorry. I'm new here.
 
  • #8
wreckemtech said:
Sorry. I'm new here.

No problem. I went through the same thing when I was just starting. Anyway, welcome to Physics Forums!

Chet
 
  • #9
thanks guys got the answer
 

1. What is "Cherry Pie Entropy"?

"Cherry Pie Entropy" is a hypothetical scenario used in thermodynamics to help explain the concept of entropy. It involves a closed system consisting of a cherry pie and its surroundings, where heat is added to the pie and its entropy increases.

2. How is the final temperature of the cherry pie determined?

The final temperature of the cherry pie is determined using the formula: ΔS = Q/T, where ΔS is the change in entropy, Q is the heat added to the pie, and T is the final temperature.

3. What is the relationship between heat and entropy in the Cherry Pie Entropy scenario?

In thermodynamics, heat and entropy are directly proportional to each other. This means that as heat is added to the cherry pie, its entropy will also increase.

4. How does the concept of entropy apply to the real world?

Entropy is a fundamental concept in thermodynamics and has many real-world applications. It helps explain why certain processes are irreversible and why energy tends to disperse and become less useful over time.

5. Can the Cherry Pie Entropy scenario be used to solve practical problems?

While the Cherry Pie Entropy scenario is a simplified example used for educational purposes, the principles and equations involved can be applied to solve practical problems in thermodynamics and other fields such as chemistry and engineering.

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