When a 9.00 g sample of metal is heated to 100.0 C, anddropped into 20.0 mLs of

Question

When a 9.00 g sample of metal is heated to 100.0 C, anddropped into 20.0 mLs of water at 25.00 C, the final temperature is28.9 C. a.) calculate the specific heat of the metal, assuming perfectinsulation and a calorimeter heat capacity of zero. b.) Estimate the atomic weight of the element and guess itsidentity. When a 9.00 g sample of metal is heated to 100.0 C, anddropped into 20.0 mLs of water at 25.00 C, the final temperature is28.9 C. a.) calculate the specific heat of the metal, assuming perfectinsulation and a calorimeter heat capacity of zero. b.) Estimate the atomic weight of the element and guess itsidentity.

Explanation / Answer

q = mc(Final temp-initial temp) q= heat evolve in the reaction m=mass c=specific heat You should be given the density of water and it's specific heat (orit could be found in your book) density of water = 1g/ml c=4.184 joule/gram °C the heat that the metal lost will be equal to negative the heatthat the water gains so -mc(Tf-Ti)=mc(Tf-Ti) g = density * ml g of h2o = 1 * 20 g of h2o = 20 Tf for both the metal and water is 28.9C now you have mass of metal = 9g Tf metal = 28.9C Ti metal = 100C mass water = 20g c of water = 4.186 joule/gram °C Tf water = 28.9 Ti water = 25C now plug into mc(Tf-Ti)=mc(Tf-Ti) to find c of metal. c metal = .51 joule/gram °C there is a formula for finding atomic weight. c * mm = a constant c   x   mm = 25 J/mol C mm = 49 g/mol ---> closest to titanium

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