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Thermochemistry, heat calculation.

  1. Jan 10, 2012 #1
    1. The problem statement, all variables and given/known data

    When 10.00 mL of a solution of strong acid is mixed with 100.00 mL of a solution of weak base in a coffee-cup calorimeter, the temperature rises from 22.8 oC to 26.8 oC. Determine q for the acid-base reaction, assuming that the liquids have densities of 1.00 g/mL and the same heat capacities as pure water.

    2. Relevant equations

    total volume V=110.0mL
    change in temperature delta T = 4
    mass m = 110.0g
    heat capacity for water C=75.291J/(mol[itex]\circ[/itex]C

    q = nC[itex]\Delta[/itex]T
    q = mC[itex]\Delta[/itex]T

    3. The attempt at a solution

    so I used the formula

    q = mC[itex]\Delta[/itex]T

    because with the information given i couldn't find the number of moles and because only the molar heat capacity is given to use i had to find the mass based heat capacity and I found it to be C = 4181.3J/KgK

    and then found q = 460J

    which then the computer told me it was wrong.

    It seemed odd that i had to go find the heat capacity rather than use the one provided. So I feel already there's my first mistake. But i know the formula is right so I can't see why it wouldn't work. However clearly my chemistry isn't that strong so perhaps I'm missing something. Any guidance would be appreciated! Thanks
  2. jcsd
  3. Jan 18, 2012 #2


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    Gold Member

    How did you get 460 J for q=mcΔT when m=0.11 kg, ΔT=4 and c=4181.3 J/kgK?

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