Question

An insulated calorimeter at 5.00ºC with a volume of 20.60 L and having a heat capacity of 25.81 J ºC-1 is filled with air at 5.00ºC. A silver spoon with a mass of 25.04 g at a temperature of 100.00ºC is placed into the container. After a short while, the temperature of the container, spoon and air was 14.75ºC. (3a) If the specific heat of silver is 0.2360 J g-1 ºC-1, calculate the heat capacity (J ºC-1) of the air in the container. (3b) If the density of air is 1.250 g L-1, calculate the specific heat of air in J g-1 ºC-1

Answer 1

Calorimter data (+ air)

Tc = 5°C; V = 20.6 L; C = 25.81 J/°C

Spoon:

m = 25.04 g; T = 100°C;

Tf = 14.75°C

Calculate specific heat capacity via:

Heat balance

-Qlost = Qgain

-Qspoon = Qcalorimter + Qair

Substitute (Q =MC(TF-Ti)

-msilver *Cpsiolver(Tfinal - Tsilver) = Ccal*(Tfinal - Tcal) + Cair*(Tf-Tcal)

substitute

-25.04*0.2360*(14.75-100) = 25.81*(14.75-5) + Cair*(14.75-5)

Solve for Cair

-25.04*0.2360*(14.75-100) - 25.81*(14.75-5) =  Cair*(14.75-5)

Cair = 252.132 / (14.75-5) = 25.859 J/°C

b)

find specific heat of air

Assume:

mass of air = Density of air * Volume of air = 1.25 g/L * 20.60 L = 25.75 g

Specific heat of air = Cair/mass of air= 25.859 /25.75 = 1.00423 J/gC