Question

1 g of α -D-glucose is burned in the adiabatic bomb calorimeter. The bomb is surrounded by 2.500 L of H₂O at 24.030°C. The bomb is made of steel and weighs 14.05 kg. Specific heats at constant pressure of water and steel at 24°C are 4.180 and 0.450 J/(g °C), respectively. The density of water at 24°C is 0.9973 g/cm³. Assuming the heat capacity of the chemicals in the bomb is negligible compared with the heat capacity of the bomb and surrounding water, find the final temperature of the system. Neglect the temperature dependence of c_P. Neglect the changes in thermodynamic functions that occur when the reactants and products are brought from their standard
states to those that occur in the calorimeter.

T_{final= ? °C .}

Answer 1

V = 2.5 L of wter

T = 24.03 C

m = 14.05 kg of calorimeter (steel)

find T final....

Hcombustion = -2805 kJ/mol

1 mol = 180

mol = mass/MW = 1/180 = 0.00555555 mol

so...

Q = Hrxn*n =-2805 * 0.00555555 = -15.58kJ = -15580 J

then...

Qlost = -Qgain

15580 = Qgain

Qgained = Qwater + Qsteel

Qgain = m*c*dT + m*Ciron*dT

15580 = (2500)(4.814)(Tf-24.03) + (14050)(0.450)(Tf-24)

calculate Tf

15580 = 10460*Tf - 24.03*10460 + 6525*Tf - 24*6525

Tf(10460+6525) = 15580 +  24.03*10460 + 24*6525

Tf = (423533.8)/(10460+6525) = 24.93575 °C

Tf = 24.93575 °C