Posted by **J** 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 -
**Devron**, 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 -
**DrBob222**, 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 -
**Cesar**, Thursday, November 20, 2014 at 1:16am
29.75

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