Entropy Change in a System with a Man Drinking Water

In summary, the man with a temperature of 310.15 K and a mass of 70 kg drinks 0.4536 kg of water at 275 K. The entropy of the man and water system increase by -0.00239 kg/K.
  • #1
scharry03
12
1

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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  • #2
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] ?
 
  • #3
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
 
  • #4
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?
 
  • #5
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
 

FAQ: Entropy Change in a System with a Man Drinking Water

What is entropy?

Entropy is a measure of the disorder or randomness within a system. It is a concept commonly used in thermodynamics to describe the state of a system.

How does entropy affect drinking water?

In terms of drinking water, entropy plays a role in the overall quality and purity of the water. As water moves through pipes and containers, it can pick up impurities and contaminants which increase its entropy. This is why it is important to filter and treat drinking water to maintain its quality.

Can entropy be reversed in drinking water?

Entropy can be reversed to a certain extent through processes such as purification and filtration. These methods remove impurities and restore the water to a lower entropy state. However, it is important to note that complete reversal of entropy is not possible, as it is a natural process that occurs in all systems.

How does temperature affect entropy in drinking water?

Temperature has a direct impact on the entropy of drinking water. As the temperature increases, the molecules in the water move faster and become more disordered, increasing the overall entropy of the system. This is why it is important to store drinking water at appropriate temperatures to maintain its quality.

Is there a maximum level of entropy that drinking water can reach?

There is no maximum level of entropy that drinking water can reach. However, as the entropy increases, the water becomes less and less suitable for consumption. This is why it is important to monitor and maintain the entropy levels in drinking water to ensure its safety and quality.

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