Chemistry
posted by J .
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)?

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=46.3ºC78.3º C=32.0ºC
m2=5.00g
c2=?
∆T2=46.3ºC23.8ºC=22.5
Solve for c2,
c2=m1c1∆T1/m2∆T2 
This must be a made up problem. I don't know that I've seen anything with a specific heat of this magnitude.

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