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wunderkind
Jun23-04, 11:07 PM
Thanks so very much with the help you gave me. I am truly grateful.
By the way, here's another weird problem:
A sample of with a mass of 250.g is heated to 130. degrees celcius, and dropped into 425g of H2O at 26.0 degrees C. The final temp. of the H2O is 38.4 degrees C. What is the specific heat capacity of the metal if 1.2 (10 exponent 3) Joules is lost to the enviroment.
Once again, I am clueless!

Gokul43201
Jun24-04, 09:33 AM
Heat lost by sample = heat gained by water + heat gained by surroundings

jatin9_99
Jun30-04, 08:40 PM
specific heat = (4.18 x 425g of water x (38.4-26)+1.2)/(250x (130-38.4))