Explain the temperature changes when anhydrous sodium carbonate is added to water and calculate its heat of solution.

The initial temperature ($T_{initial}$) is $20.5^\circ C$ (at Time = 0 s).
The temperature stabilizes at $27^\circ C$ (from Time = 7 s to 9 s).

$$\Delta T = T_{final} - T_{initial} = 27^\circ C - 20.5^\circ C = 6.5^\circ C = 6.5 \text{ K}$$

Mass of anhydrous $Na_2CO_3 = 8.5 \text{ g}$.
Volume of distilled water = $50 \text{ cm}^3$.
Assuming the density of water is $1 \text{ g cm}^{-3}$, the mass of water is $50 \text{ g}$.

$$m_{solution} = \text{Mass of } Na_2CO_3 + \text{Mass of water} = 8.5 \text{ g} + 50 \text{ g} = 58.5 \text{ g}$$

The specific heat capacity of the solution ($c$) is $4.2 \text{ J g}^{-1} \text{K}^{-1}$.

$$Q_{solution} = m_{solution} \times c \times \Delta T$$ $$Q_{solution} = 58.5 \text{ g} \times 4.2 \text{ J g}^{-1} \text{K}^{-1} \times 6.5 \text{ K}$$ $$Q_{solution} = 1597.05 \text{ J}$$

Molar mass of $Na_2CO_3$:

$$(2 \times 22.99 \text{ g/mol for Na}) + (1 \times 12.01 \text{ g/mol for C}) + (3 \times 16.00 \text{ g/mol for O}) = 105.99 \text{ g/mol}$$ $$\text{Moles of } Na_2CO_3 = \frac{\text{Mass}}{\text{Molar mass}} = \frac{8.5 \text{ g}}{105.99 \text{ g/mol}} \approx 0.080196 \text{ mol}$$

Since the temperature increased, the dissolution is an exothermic process, so $\Delta H_{sol}$ will be negative.

$$\Delta H_{sol} = -\frac{Q_{solution}}{\text{Moles of } Na_2CO_3}$$ $$\Delta H_{sol} = -\frac{1597.05 \text{ J}}{0.080196 \text{ mol}}$$ $$\Delta H_{sol} \approx -199140 \text{ J/mol}$$

Converting to kilojoules per mole:

$$\Delta H_{sol} \approx -199.14 \text{ kJ/mol}$$

Rounding to two significant figures (based on $\Delta T$ and mass of $Na_2CO_3$):

$$\Delta H_{sol} \approx \boxed{-200 \text{ kJ/mol}}$$