Question

5.17 grams of CaCl 2 was added to 85.0 cm 3 of water in a Styrofoam cup calorimeter. The temperature of the water changed from 21.4°C to 27.5 °C.
Calculate the molar q value (q mol) for dissolving 5.17 grams of CaCl 2 in 85.0 cm 3 of water.
Assume the heat capacity of the system is due only to the water in the system. The specific heat of water is 4.18 J/(g∙°C).

Answer 1

First we should calculate the amount in moles of CaCl₂ inserted into the water calorimeter:

M(CaCl₂) = 111 g/mol;

$n(CaCl_2)=\frac{m(CaCl_2)}{M(CaCl_2)}=\frac{5.17g}{111g/mol}=0.0466mol$

The temperature in the calorimeter was changed on 27.5C - 21.4C = 6.1C. The mass of water in the calorimeter is 85.0 g. Thus, the amount of heat released in the calorimeter during the dissolution of CaCl₂ was equal to:

Q(CaCl₂) = C(H₂O)m(H₂O)(t₂ - t₁) = 4.18 Jg^-1C^-1 85.0g 6.1C = 2167J

Now, it is the right time to found the specific molar heat of CaCl₂ dissolution:

$Q_{spec}=\frac{Q(CaCl_2)}{n(CaCl_2)}=\frac{2167J}{0.0466mol}=46.5Jmol^{-1}$