Entropy Change in a System with a Man Drinking Water

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Homework Help Overview

The problem involves a man with a constant body temperature drinking water at a lower temperature, and the goal is to find the entropy change of the entire system. The subject area includes thermodynamics, specifically the concepts of entropy and heat transfer.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the application of two different equations for calculating entropy change, questioning when to use each. There is confusion regarding the implications of the man's body temperature remaining constant and how that affects the calculations. Some participants suggest re-evaluating the problem by considering the heat transfer and the final temperature of the system.

Discussion Status

The discussion is active, with participants exploring different interpretations of the problem and the equations involved. Some guidance has been offered regarding the treatment of the man as a finite reservoir and the implications of limits in the context of the equations. There is no explicit consensus yet on the approach to take.

Contextual Notes

Participants note the assumption that the man's temperature does not change and the implications of this assumption on the calculations. There is also mention of a potential heat balance approach that could be considered to find the final equilibrated temperature.

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Homework Statement


Man with a temperature of 310.15 K and a mass of 70 kg drinks 0.4536 kg of water at 275 K. Ignoring the temperature change of the man from the water intake (assume human body is a reservoir always at same temperature), find entropy increase of entire system.

Homework Equations


dS = - absvalue(Q)/Tbody
deltaS = mcln(Tfinal/Tinitial)

The Attempt at a Solution


I'd tried solving only using the second equation on here, but the answer only used that equation for the deltaS of the water. The natural log of the body would be 0, since T is constant, so I thought there wouldn't be an enthalpy change for the body. I don't understand under what circumstances the first equation would be used instead of the second equation. Also, the numbers plugged into the first equation for the deltaS(body) are -m(water)(1) [(Tbody-Twater)/Tbody] which is nothing like what the first equation asks for, so I am very confused.
 
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scharry03 said:

Homework Statement


Man with a temperature of 310.15 K and a mass of 70 kg drinks 0.4536 kg of water at 275 K. Ignoring the temperature change of the man from the water intake (assume human body is a reservoir always at same temperature), find entropy increase of entire system.

Homework Equations


dS = - absvalue(Q)/Tbody
deltaS = mcln(Tfinal/Tinitial)

The Attempt at a Solution


I'd tried solving only using the second equation on here, but the answer only used that equation for the deltaS of the water. The natural log of the body would be 0, since T is constant, so I thought there wouldn't be an enthalpy change for the body. I don't understand under what circumstances the first equation would be used instead of the second equation. Also, the numbers plugged into the first equation for the deltaS(body) are -m(water)(1) [(Tbody-Twater)/Tbody] which is nothing like what the first equation asks for, so I am very confused.
Does it mean another way to find deltaS(hot) is to do m(cold) x c(cold) x [(T(hot)-T(cold)/Thot] ?
 
For the case of the constant temperature reservoir, this is a limiting case taken in the limit as its mass times its heat capacity becomes very large. So the second equation does actually apply to this situation, but only in the limit. Try doing the problem again, but instead, treat the reservoir as having a finite product of mass times heat capacity. What is the equation for the final temperature in this situation? Based on this result, what is the equation for the change in entropy of the water, the reservoir, and the total? Now, take the limit of the equations as the mass times the heat capacity of the reservoir becomes very large. See what you get.

Chet
 
I've never framed these questions in terms of limits as you are doing, so I don't really understand what you mean. I had a chance to ask my professor about this and he said the first equation is used in something whose temperature isn't changing, and the second is used in the case that it is. How would you rephrase this in terms of limits, so that I can understand that method of thinking?
 
scharry03 said:
I've never framed these questions in terms of limits as you are doing, so I don't really understand what you mean. I had a chance to ask my professor about this and he said the first equation is used in something whose temperature isn't changing, and the second is used in the case that it is. How would you rephrase this in terms of limits, so that I can understand that method of thinking?
I can help you with this if you are willing to work with me. Instead of ignoring the temperature change of the man, we will include it.

Let:

Mm=mass of man
Cm=heat capacity of man
Mw= mass of water
Cw=heat capacity of water
Th=starting temperature of man (e.g., 310)
Tw=starting temperature of water (e.g., 275)
T = final equilibrated temperature of man and water

From a heat balance (algegraically), what is T in terms of the other parameters?
How much heat is transferred from the man to the water to achieve this temperature change?
Using the second equation, what is the entropy change of the man (algebraically)?
Using the second equation, what is the entropy change of the water (algebraically)?

Let's stop here and see what you come up with. Then we can take the limits as Mm becomes large.

Chet
 

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