Question

Magnesium sulfate is often used in first-aid hot packs, giving off heat when dissolved in water. When \(2.00 \mathrm{~g}\) of \(\mathrm{MgSO}_{4}\) dissolves in \(15.0 \mathrm{~mL}\) of water \((d=1.00 \mathrm{~g} / \mathrm{mL})\) at \(25^{\circ} \mathrm{C}\) in a coffee-cup calorimeter, \(1.51 \mathrm{~kJ}\) of heat is evolved. (a) Write a balanced equation for the solution process. (b) Is the process exothermic? (c) What is \(q_{\mathrm{H}_{2} \mathrm{O}} ?\) (d) What is the final temperature of the solution? (Specific heat of water is \(4.18 \mathrm{~J} / \mathrm{g} \cdot{ }^{\circ} \mathrm{C}\).) (e) What are the initial and final temperatures in \({ }^{\circ} \mathrm{F} ?\)

Short answer

Answer: The final temperature of the magnesium sulfate solution is approximately \(49.00^{\circ} \mathrm{C}\) (Celsius) and \(120^{\circ} \mathrm{F}\) (Fahrenheit).

Step-by-step solution

Write a balanced equation for the solution process.

The balanced equation for the dissolution of magnesium sulfate in water is: \(\mathrm{MgSO}_{4} (s) \rightarrow \mathrm{Mg}^{2+} (aq) + \mathrm{SO}_{4}^{2-} (aq)\) This equation shows that one molecule of solid magnesium sulfate (\(\mathrm{MgSO}_{4}\)) dissolves in water to form one magnesium ion (\(\mathrm{Mg}^{2+}\)) and one sulfate ion (\(\mathrm{SO}_{4}^{2-}\)). #b)

Determine if the process is exothermic.

Since the dissolution of magnesium sulfate in water releases heat (\((1.51 \mathrm{~kJ}\) of heat is evolved), the process is exothermic. #c)

Calculate the heat transfer, \(q_{\mathrm{H}_{2} \mathrm{O}}\).

Heat evolved from the dissolution process is equal to the heat absorbed by water. So, \(q_{\mathrm{H}_{2} \mathrm{O}}=1.51 \mathrm{~kJ}=1510 \mathrm{~J}\) #d)

Calculate the final temperature of the solution.

To find the final temperature, we need to use the formula: \(q_{\mathrm{H}_{2} \mathrm{O}} = m_{\mathrm{H}_{2} \mathrm{O}} \times C_{p,\mathrm{H}_{2} \mathrm{O}} \times \Delta T\) where \(q_{\mathrm{H}_{2} \mathrm{O}}=1510 \mathrm{~J}\), \(m_{\mathrm{H}_{2} \mathrm{O}}=15.0 \mathrm{~g}\) (mass of the water), \(C_{p,\mathrm{H}_{2} \mathrm{O}}=4.18 \mathrm{~J} / (\mathrm{g} \cdot { }^{\circ} \mathrm{C}\)) (specific heat capacity of water), and \(\Delta T = T_\text{final} - T_\text{initial}\) is the temperature change. We need to solve for \(T_\text{final}\): \(1510 \mathrm{~J} = (15.0 \mathrm{~g}) \times (4.18 \mathrm{~J} / (\mathrm{g} \cdot {^{\circ} } \mathrm{C})) \times (T_\text{final} - 25^{\circ} \mathrm{C})\) \(T_\text{final} - 25^{\circ} \mathrm{C} = \frac{1510 \mathrm{~J}}{(15.0 \mathrm{~g}) \times (4.18 \mathrm{~J} / (\mathrm{g} \cdot {^{\circ} } \mathrm{C}))}\) \(T_\text{final} - 25^{\circ} \mathrm{C} \approx 24.00^{\circ} \mathrm{C}\) \(T_\text{final} \approx 49.00^{\circ} \mathrm{C}\) So, the final temperature of the solution is approximately \(49.00^{\circ} \mathrm{C}\). #e)

Calculate the initial and final temperatures in \({ }^{\circ} \mathrm{F}\).

To convert temperatures from Celsius to Fahrenheit, we use the formula: \(T_\mathrm{F}=\frac{9}{5} \times T_\mathrm{C}+32\) Initial temperature in Fahrenheit: \(T_\text{initial} = \frac{9}{5} \times (25^{\circ} \mathrm{C}) + 32 \approx 77^{\circ} \mathrm{F}\) Final temperature in Fahrenheit: \(T_\text{final} = \frac{9}{5} \times (49.00^{\circ} \mathrm{C})+32 \approx 120^{\circ} \mathrm{F}\) So, the initial and final temperatures of the solution are approximately \(77^{\circ} \mathrm{F}\) and \(120^{\circ} \mathrm{F}\), respectively.