1. Not finding help here? Sign up for a free 30min 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!

Equilibrium Temperature of water in cup

  1. Sep 15, 2007 #1
    1. The problem statement, all variables and given/known data

    A 37 g ice cube at 0.0°C is added to 110 g of water in a 62 g iron cup. The cup and the water have an initial temperature of 40°C.
    (a) Find the equilibrium temperature of the cup and its contents.


    2. Relevant equations
    I think a relevant equation is (m Lsubf) + (mc delta T) + (mc delta T)


    3. The attempt at a solution

    I am completely lost, I started out (.110kg)(4.186 j/ kg c) (T-40) + (.037)(2090 j/ kg c)(T-0) + (.062) (448 j/kg c) (T-40) , I think I have most the units correct, I just don't know the correct equation for the problem or maybe what to do with all the units. Help please, am I using the correct equation? am I missing units? What is L sub f ?

    I added the like units and mutiplied them then divided by T value to get T:
    (.209kg)(2542.186 j/kg c)/80 and got T= 6.64 C. This came up as incorrect and I was wondering if anyone could tell me what I did wrong?
     
    Last edited: Sep 15, 2007
  2. jcsd
  3. Sep 15, 2007 #2

    learningphysics

    User Avatar
    Homework Helper

    Your equation looks right:

    (.110kg)(4.186 j/ kg c) (T-40) + (.037)(2090 j/ kg c)(T-0) + (.062) (448 j/kg c) (T-40)


    = 0

    solve for T.
     
  4. Sep 15, 2007 #3
    I added the like units and mutiplied them then divided by T value to get T:
    (.209kg)(2542.186 j/kg c)/80 and got T= 6.64 C. This came up as incorrect and I was wondering if anyone could tell me what I did wrong?
     
  5. Sep 16, 2007 #4

    dynamicsolo

    User Avatar
    Homework Helper

    I think you have converted some of your heat capacities and not others: iron would be 448 j/kg c, but then water must be 4186 j/kg c .

    I take it that the 2090 j/kg c is what you are using for the latent heat of fusion (or melting of ice). The process of melting the ice is a phase transition which occurs at constant temperature 0 C, so there shouldn't be any temperature factor there. But also, the latent heat should be 334,000 j/kg , no, since it's 80 cal/gm? (You don't need a warming term for the ice, since its initial temperature is given as 0 C.)

    You are also missing a term: the 37 g of meltwater from the ice will be warming up from 0 C, so you need a (.037)(4186 j/kg)(T - 0) to account for that.

    If you set up the calorimetry equation with all the terms on one side, don't forget that the ice is warming up, but the water and iron cup are *cooling down*. The temperaure change factors would have to be written as (T final - T initial).

    You'll need to rewrite your equation in this light and try solving it again.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Equilibrium Temperature of water in cup
  1. PSI of water in a cup (Replies: 8)

Loading...