An unknown mass of copper (s=0.385) at 400 degrees celsius is placed into a cup containing 30 g of water at 18 degrees celsius. If the final system of water and copper reaches a temperature of 58 degrees celsius, what mass of copper was used?
+37146 Yes, I do.
The heat the copper loses is GAINED by the water
specific heat of water = 4.186 j/(gm deltaT)
Let m= mass of copper
m (.385) (400-58) = 30(4.186)(58-18) solve for m
m= 30(4.186)(40) / ((.385)(342))
m= 38.15 gm
Don't have a formula in your book relating the mass, specific heat, temperature of an element or material. Must have a chapter on it.
It would help if you gave some sort of a formula. If you do, make sure that you explain what various letters, symbols, variables stand for. For example, in your question you have(s=.385)? Does "s' stand for specific heat or what?......etc. Remember, this is a math forum and there aren't many people here who are very good at Chemistry, with a possible exception of "Alan", who is only online for a few hours a day.
