1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Entropy - Drinking Water

  1. Jul 13, 2015 #1
    1. The problem statement, all variables and given/known data
    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.

    2. Relevant equations
    dS = - absvalue(Q)/Tbody
    deltaS = mcln(Tfinal/Tinitial)

    3. 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.
  2. jcsd
  3. Jul 13, 2015 #2
    Does it mean another way to find deltaS(hot) is to do m(cold) x c(cold) x [(T(hot)-T(cold)/Thot] ?
  4. Jul 14, 2015 #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.

  5. Jul 14, 2015 #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?
  6. Jul 15, 2015 #5
    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.


    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.

Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted