# How to calculate the temperature change when CaCl2, NaCl, and water added together at an initial temperature of 298 K?

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?

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