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(0.04)c(33.7-23.5)+(0.08)c(33.7-39.6) Find C

  1. Dec 1, 2015 #1
    Sorry if the title is against the rules or anything, I just wanted to be specific as possible :P
    1. The problem statement, all variables and given/known data

    V1(Saltwater) = 40g = 0.04kg
    V2(Hot-Saltwater) = 80g = 0.08kg
    Ti(Saltwater) = 23.5oc
    Ti(Hot-Saltwater) = 39.6oc
    Tf = 33.7c

    Question: A cup of 40g saltwater is at 23.5c & A cup of 80g saltwater is at 39.6c, Find the specific heat capacity of salt water. **I measured these myself because it was a mini-lab so the specific heat capacity might not be exactly 3.99x103J.

    2. Relevant equations

    I asked my grade 11 physics teacher and this could be solved by using QLOST + QGAINED and Q=mcΔT. I asked my friends and they couldn't figure it out too ;-;

    3. The attempt at a solution
    I tried using everything and did anything I could do ....but I still couldn't get the answer of 3.99x103J. I can easily figure it out if I was it was 2 different substances and at least one heat capacity, but finding only c because they are both salt water confuses me.
    m1cΔT1 + m1cΔT1
    (0.04)c(33.7-23.5)+(0.08)c(33.7-39.6)
     
  2. jcsd
  3. Dec 1, 2015 #2

    mfb

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    Staff: Mentor

    Without a reference (known input heat or something with a known heat capactiance), the whole approach cannot work. You are comparing salt water with salt water.
     
  4. Dec 1, 2015 #3
    Oh alright, yeah I just wanted to make sure it wasn't possible. My teacher does this thing when he tricks us to check if we actually do our work... it's really annoying but he does teach us why it's impossible and stuff like that. Thanks for your help and time :)
     
  5. Dec 2, 2015 #4

    mfb

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    Staff: Mentor

    You can check the numbers for consistency: Check if the mixing temperature is right.
     
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