The enthalpy of combustion of lactose, C12H22O11, is - 5652 kJ/mol. A 2.50 g sample of lactose was burned in a calorimeter that contained 1350 g of water. The heat capacity of the calorimeter is 1630 J/oC, and the initial temperature was 24.58oC. What was the final temperature (oC) ?

30.25°C

Explanation:

The calorimeter is an equipment used to measure the combustion enthalpy of a substance. The heat loss in the reaction is used to heat the water and the equipment. By the conservation of energy:

Qcombustion + Qcalorimeter + Qwater = 0

Because there is no phase change:

Qcalorimeter = C*ΔT, where C is the heat capacity, and ΔT the variation in temperature (final - initial)

Qwater = m*c*ΔT, where m is the mass, and c is the specific heat (4.184 J/g°C).

The molar mass of lactose is 342.3 g/mol, so the number of moles in 2.50 g is:

n = mass/molar mass

n = 2.50/342.3

n = 0.0073 mol

Qcombustion = - 5652 kJ/mol * 0.0073 mol

Qcombustion = - 41.28 kJ

Qcombustion = - 41280 J

Thus,

-41280 + 1630 * (T - 24.58) + 1350*4.184 * (T - 24.58) = 0

(T - 24.58) * (1630 + 5648.4) = 41280

7278.4 (T - 24.58) = 41280

T - 24.58 = 5.67

T = 30.25°C