In the laboratory a \"coffee cup\" calorimeter, or constant pressure calorimeter

Question

In the laboratory a "coffee cup" calorimeter, or constant pressure calorimeter, is frequently used to determine the specific heat of a solid, or to measure the energy of a solution phase reaction. A chunk of titanium weighing 19.89 grams and originally at 97.85 oC is dropped into an insulated cup containing 84.26 grams of water at 20.59 oC. The heat capacity of the calorimeter (sometimes referred to as the calorimeter constant) was determined in a separate experiment to be 1.57 JVOC. Water Using the accepted value for the specific heat of titanium (See the References tool), calculate the final temperature of the sample water. Assume that no heat is lost to the surroundings. final Stirring rod

Explanation / Answer

m = 19.89 g of Ti

Ti = 97.84°C

m water = X

20.59 °C

Qcal = 1.57 J/°C

find T final of water

Qtitanium + Qwater + Qcal = 0

Qti = mti*Cti*(Tf-Tti)

Qw = mw*Cw*(Tf-Tw)

Qcal = Ccal*(Tf-Tw)

substitute all

Qtitanium + Qwater + Qcal = 0

mti*Cti*(Tf-Tti) + mw*Cw*(Tf-Tw) + Ccal*(Tf-Tw) = 0

19.89*0.54*(Tf-97.85) + 84.26*4.184*(Tf-20.59) + 1.57*(Tf-20.59) = 0

solve for Tf

10.74*Tf -97.85*10.74 + 352.5*Tf - 20.6*352.5 + 1.57*Tf - 1.57*20.59 = 0

Tf (10.74 + 352.5 + 1.57) =  1.57*20.59 + 20.6*352.5+97.85*10.74

Tf = ( 1.57*20.59 + 20.6*352.5+97.85*10.74 ) / (10.74 + 352.5 + 1.57)

Tf = 22.87°C

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