Oh no! Billy forgot that his calorimeter should have 2 coffee cups to contain all the heat. If Billy dropped a 25.0 g piece of solid iron (Cs= 0.449 J/g°C) at 398 K in a coffee cup containing 25.0 mL of water at 298K and because of his mistake, 21% of the heat that should be transferredfrom the iron to the wateris lost to the atmosphere, calculate the final temperature of the iron water mixture?

Explanation:

Let the final temperature be T .

heat lost by iron = mass x specific heat x decrease in temperature

= 25 x .449 x ( 398 - T  )

Heat applied to increase the temperature of water

= 25 x .449 x ( 398- T  )  x .79  [ 21 % is lost ]

= 8.86775 ( 398 - T  )

heat gained by water = 25 x 4.2 x ( T - 298 )

Both of them are equal

8.86775 ( 398 - T  )  = 25 x 4.2 x ( T - 298 )

8.86775 ( 398 - T  )  = 105 x ( T - 298 )

105 T + 8.86775 T = 105 x 298 + 398 x 8.86775

113.86775 T = 31290 + 3529.36

113.86775 T = 34819.36

T = 305.78 K .

frequency, z, and the collision density, z, in carbon dioxide (co2), r 180 pm at 30 °c and 120 kpa

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