Titanium and colorimeter?

Question: An experiment is preformed where a 72.6g piece of titanium metal is heated to 98.2C in a boiling water bath. the titanium is then quickly transferred into 100ml of calorimeter water at 23.5C. the final temp of the mixture in the calorimeter reaches 29.4C a)list two assumptions that you would have to make in order to calculate the specific heat capacity of titanium b)Calculate the experimental specific heat capacity of the titanium sample

Answer: a. There are a bunch of assumptions that are required. The most obvious one is that the titanium piece did not cool during transfer. Another one is that the heat given off by titanium is completely absorbed by the water in the calorimeter. Also, you must assume that the titanium sample is 98.2C throughout, not just on the surface. Also, that no water evaporated from the calorimeter upon insertion of the titanium sample. Also, assume that the heat capacity (of water and of titanium) does not depend on temperature (it actually does but not very much). You can pick any two, but the first two are the most appropriate for you. b. Find the mass of water 100mL * 0.997 g/mL (density of water @23.5C) = 99.7 g. for water: Q=mC(deltaT)= Q=99.7g*4.184J/gC* (29.4-23.5)C= 2461J for titanium: Q=mC(deltaT) or C=Q/m/deltaT Q is the same as before (assumption 2): C = 2461J/72.6g/(98.2-29.4C)= 0.4927 (round off to 0.493 J/gC) (the true value is 0.523, so this answer is 5% off, not that it matters for this problem)

Related Questions

Related Items