Question

Consider the dissolution of$CaC{l}_2$ :

$CaC{l}_2\left(s\right)\to C{a}^{2+}\left(aq\right)+2C{l}_2\left(aq\right)∆H=-81.5kJ$

An 11.0 g sample of $CaC{l}_2$is dissolved in 125 g of water, with both substances at $25.0°C$. Calculate the final temperature of the solution assuming no heat loss to the surroundings and assuming the solution has a specific heat capacity of $4.18J/°Cg$.

Short answer

The final temperature of the solution is $41.4°C$

Step-by-step solution

To calculate heat produced

$∆H=\frac{q}{n}$ here, $∆H=81.5KJ/mol$

Mass of $CaC{l}_2=15.0g$

Molar mass of $CaC{l}_2=110.98g/mol$

$\begin{array}{rcl}MolesofCaC{l}_2& =& \frac{MassofCaC{l}_2}{MolecularMassofCaC{l}_2}\\ & =& \frac{15.0g}{110.98g/mol}\\ & =& 0.135mol\\ & & \end{array}$

Put the values in the formula,$∆H=\frac{q}{n}$

We get, $q=11.0KJ$

 To calculate the final temperature of the solution

Formula: $q=mC(T_2-T_1)$

Here, q = heat produced

Mass of solution = m

Specific heat capacity

T₁=initial temperature

T₂ = final temperature = ?

By using the formula;

$q=mC(T_2-T_1)$

$$\begin{gathered} 11000J=160g\times 4.18J/°C-g\times (T_2-25.0) \\ {T}_2=41.4°C \end{gathered}$$