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Thermal dynamics

  1. Nov 30, 2005 #1
    A 330-g glass mug at 16C is filled with 275 milliliters of water at 91C. Assuming no losses to the external environment, what is the final temperature of the mug?

    Im having a bit of trouble with this problem. Since we're assuming no losses, delta q=0.

    So, delta q= delta q(mug) + delta q (water) = 0

    and delta q(mug) = mass(mug)*specific heat(mug)*delta T(mug)
    and delta q(water) = mass(water) *specific heat(water)*delta T(mug)

    so put it all together and i get
    0=.330kg*840J/kgK*delta T(mug) + .275kg*4186J/kgK*delta T(water)

    But where do i go from here? Is the change in temperature for botht he mug and the water going to be the same? in which case i could factor out a delta T, but then that still leaves me stuck.

    Does the change in temperature of the water even matter at all?

    Theres got to be a way to do this that im not seeing...

    Thanks
     
  2. jcsd
  3. Nov 30, 2005 #2

    Chi Meson

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    It is the final temperature that is the same for both. They both have their own initial temperatures, right? "Delta T" means "final T minus initial T." So substitute ([tex]T_f - T_o[/tex]) where the delta T is.
     
  4. Nov 30, 2005 #3
    sweet thanks. that pointed me in the right direction i do believe

    i substituted (Tf-To) for delta T, and solved for Tf coming up with a final temperature of 76.46 degrees celclius. that sound right?

    thanks again for the help
     
  5. Nov 30, 2005 #4

    Chi Meson

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    okie dokie. I don't have my calculator handy, but that seems aobut right.
     
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