Science

In the calibration of a calorimeter, an electrical resistsnce heater supplies 80J of heat and a temperature increase of 1.20°C is observed. What is the initial temperature of 8.50g of (C8H18) with a molar heat capacity of 39.36J/molK after heating in the same calorimeter? The initial temp of the calorimeter is 28.0°C and the final temperature of the system is 50.0°C.

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asked by Kai
  1. Q=mc∆t

    Q=80J

    80J/1.20K=mc

    mc∆T=mc∆T

    (80J/1.20K)(22K)=(39.36J/molK)*(8.50g*mol/114.23 g)(323.15K-Ti)

    ****114.23g=molecular weight of C8H18


    Solve for Ti


    Double √√ my thinking!!!!!

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  2. What does mc means im quite confused sir

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    posted by Kathy
  3. where do 22K come from?

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  4. My answer is almost 800K

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    posted by Kai
  5. Mc=specific heat of the calorimeter

    m=moles
    c=specific heat
    The initial info given in the problem is used so you can determine the specific heat of the calorimeter.

    When the water and the calorimeter are heated together, they both will gain the same amount of heat, so you can equate the equations.

    You know the specific heat of the calorimeter and the ∆T of it as well.

    You know the specific heat of the water, the final temp, mole ratio, and the specific heat--you just don't know the initial temp.


    I did the thinking not the math, but if you are getting funky numbers then the author of the problem did not check numbers and this is not a real chem problem that a chemist would make. And looking at it, you should return a negative Kelvin number, which means that the water wasn't water but ice. A totally made up problem because the numbers don't add up.




    Best,


    P.S.

    What makes you think that I am a male????

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  6. You may want to repost the problem and see what Bob222 or Pursley think......

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  7. Well, I usually call the people here sir bc I think most of the tutors here are male.

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    posted by Kai
  8. Yeah same

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    posted by Kathy
  9. This is just a hunch, but I think the molar heat capacity given as 39.36J/mol•K should be 39.36kJ/mol•K

    When I performed the calculations after he correction, and make sure that the units for (80J/1.20K)(22K)= kJ/mol•K, I come.up.with an answer that seems believable. There is a typo in the question stem, either on the author 's end of yours.

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  10. And the 22K comes from the ∆T for the calorimeter. I just converted to I and did not show my work. I have a grammatical error in my explanation: the heat that the calorimeter gives off is equal to the heat that the water gains, which is why you can equate the equations to each other.


    Bests

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