Solve Another Weird Problem: Calculate Specific Heat Capacity

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The discussion revolves around a thermodynamics problem involving heat transfer between a heated metal sample and water. A metal sample weighing 250 grams is heated to 130 degrees Celsius and then placed in 425 grams of water at 26 degrees Celsius, resulting in a final water temperature of 38.4 degrees Celsius. The problem requires calculating the specific heat capacity of the metal, taking into account that 1.2 x 10^3 Joules of heat is lost to the environment. The heat lost by the metal sample is equal to the heat gained by the water plus the heat gained by the surroundings. The specific heat capacity formula used is derived from the heat transfer equation, incorporating the mass and temperature changes of both the water and the metal sample.
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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!
 
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Heat lost by sample = heat gained by water + heat gained by surroundings
 
specific heat = (4.18 x 425g of water x (38.4-26)+1.2)/(250x (130-38.4))
 

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