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Posted by on Tuesday, March 19, 2013 at 8:14pm.

A coffee cup calorimeter contains 25.0 grams water at 23.8 C
A 5.00g sample of an unknown metal at an initial temperature of 78.3 C was dropped into the calorimeter.
The final temperature of mixture was 46.3 C.

Calculate the specific heat of the metal. The specific heat of ater is 4.184 J/ (g C)?

  • Chemistry - , Tuesday, March 19, 2013 at 9:04pm

    You will need the following equation:

    q=mc∆T

    heat lost=heat gain, that is,

    m1c1∆T1=m2c2∆T2

    Where
    m1=25.0g
    c1=4.184 J/ (g C)
    ∆T1=46.3ºC-78.3º C=32.0ºC
    m2=5.00g
    c2=?
    ∆T2=46.3ºC-23.8ºC=22.5


    Solve for c2,

    c2=m1c1∆T1/m2∆T2

  • Chemistry - , Tuesday, March 19, 2013 at 9:56pm

    This must be a made up problem. I don't know that I've seen anything with a specific heat of this magnitude.

  • Chemistry - , Thursday, November 20, 2014 at 1:16am

    29.75

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