Thermochemistry, heat calculation.

  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. ehild

    ehild 12,439
    Homework Helper
    Gold Member
    2014 Award

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

    ehild
     
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