Question

A coffee-cup (constant pressure) calorimeter is used to carry out the following reaction in an unknown volume of water (where X is a hypothetical metal): X + 2 H2O → X(OH)2 + H2 In this process, the water temperature rose from 25.0 °C to 32.2 °C. If 0.00803 mol of "X" was consumed during the reaction, and the ΔH of this reaction with respect to the system is -1798 kJ mol-1 , what volume of water (in mL) was present in the calorimeter? The specific heat of water is 4.184 J g-1 °C-1

Answer 1

Quantity of heat gained by water is given by,

Q_p = heat at constant pressure =

= change in enthalpy = -1798 kJ mol^-1 = -1798 x 1000 J mol^-1 = - 1798000 J mol^-1

mass of water

C_water = Heat capacity of water = 4.184 J g^{-1 0}C^-1

temperature change of water

Thus,

-1798000 = m x 4.184 x (25.0 - 32.2)

or, m = 1798000/(4.184 x 7.2)

or, m = 59685 g

Therefore the mass of the water in the calorimeter = 59685 g

Density of the water = 1 g/mL

Therefore the volume of the water in the calorimeter = mass of the water/ density of the water

                                                                                    = 59685/1

                                                                                    = 59685 mL