A sample of aluminum (specific heat capacity = 0.89 J/goC) is heated to 115.0oC and then placed in 500 mL of water that is at 22.3oC. The final temperature of both the water and the aluminum is
28.3oC. What is the mass of the aluminum that is added to the container, assuming that all the
heat lost by the aluminum is gained by the water? Assume that the density of water is 1.0 g/mL

answer : third trial had the contaminated sample and the correct value for dhrxn should be 1000 kj/mol

explanation :

let us write down the given equations

equation 1 :

equation 2 :

equation 3 :

let us use hess's law to find out which 2 equations can be added to get the third equation.

let us reverse equation 2 and multiply it by 2. the new equation that we get is,

equation 4 :

let us add equation 1 & 4

the above equation is similar to equation 3 but has different value for dhrxn.

therefore third trial had the contaminated sample and the correct value for dhrxn should be 1000 kj/mol