0
$\begingroup$

My question: How to calculate the temperature change when adding $\pu{0.1 kg}$ of $\ce{CaCl2}$ and $\pu{0.1 kg}$ of $\ce{NaCl}$ to $\pu{1 kg}$ water with an initial temperature of $\pu{298 K}$?

I know that the $\Delta H_\mathrm{solution}$ for $\ce{CaCl2}$ in water $= \pu{-81.3 kJ/mol}$ (heat release) and that for $\ce{NaCl}$ in water $= \pu{3.88 kJ/mol}$ (heat consumption). Also: $Q = m \times C_p \times \Delta T$

Assuming that the $C_p$ of water is the same as the $C_p$ of the water/salt mixture and that it is $\pu{4.2 kJ kg-1K-1}$.

Molar weight of $\ce{CaCl2} = \pu{111 g/mol}$. So lets say that $\pu{0.1 kg}$ of $\ce{CaCl2}$ is $\pu{1 mol}$ of $\ce{CaCl2}$.

Molar weight of $\ce{NaCl} = \pu{58 g/mol}$. So I then have $\pu{1.7 mol}$ of $\ce{NaCl}$.

I know that I can calculate the temperature change when adding $\ce{CaCl2}$ to water as follows (isolating $T_\mathrm{final}$ from the formula $Q = m \times C_p \times \Delta T$):

$T_\mathrm{final} = T_\mathrm{initial} + \frac{n_\ce{CaCl2}\times \Delta H_\mathrm{solution}}{(m_\ce{CaCl2} + m_\ce{H2O})\times C_p}$

$T_\mathrm{final} = 298 + \frac{1\times 81.3}{(0.1+1)\times 4.2} = \pu{315 K}$

When I add both the salts, do I sum the enthalpy of the solutions to have a net enthalpy of solution? If so, then I think I can calculate the final temperature as follows:

$T_\mathrm{final} = 298 + \frac{(1+1.7)(81.3-3.88)}{(0.1+0.1+1)\times 4.2} = \pu{339 K}$

This cannot be correct because the temperature of the $\ce{CaCl2 + NaCl + H2O}$ mixture should be lower than the $\ce{CaCl2 + H2O}$ mixture.

What am I doing wrong?

$\endgroup$
  • $\begingroup$ Welcome to Chemistry Stackexchange. You should use mathJax for formatting as needed in this site in the future. $\endgroup$ – Mathew Mahindaratne Nov 29 '19 at 15:48

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.