Commercial cold packs consist of solid ammonium nitrate and water.NH₄NO₃ absorbs 25.69 kJ of heat per mole dissolved in water.In a coffee-cup calorimeter,5.60 g NH₄NO₃ is dissolved in 100.0 g of water at 22.0 °C.What is the final temperature of the solution? Assume that the solution has a specific heat capacity of 4.18 J/g⋅K.

Question

Quizplus · Accepted Answer

The heat absorbed by the dissolution of ammonium nitrate ($NH_4NO_3$) can be calculated using the formula $q = m \cdot c \cdot \Delta T$, where $q$ is the heat absorbed, $m$ is the mass of the solution, $c$ is the specific heat capacity, and $\Delta T$ is the change in temperature. Given that 5.60 g of $NH_4NO_3$ absorbs 25.69 kJ/mol upon dissolution and the molar mass of $NH_4NO_3$ is approximately 80 g/mol, we first find the moles of $NH_4NO_3$ dissolved: $5.60 \, \text{g} / 80 \, \text{g/mol} = 0.07 \, \text{mol}$. The heat absorbed is $0.07 \, \text{mol} \times 25.69 \, \text{kJ/mol} = 1.7983 \, \text{kJ}$ or $1798.3 \, \text{J}$. Assuming the mass of the solution is approximately the mass of the water (since the mass of the solute is relatively small), $m = 100.0 \, \text{g}$, and $c = 4.18 \, \text{J/g⋅K}$, we can solve for $\Delta T$: $1798.3 \, \text{J} = 100.0 \, \text{g} \cdot 4.18 \, \text{J/g⋅K} \cdot \Delta T$. Solving for $\Delta T$ gives $\Delta T = 1798.3 \, \text{J} / (100.0 \, \text{g} \cdot 4.18 \, \text{J/g⋅K}) = 4.3 \, \text{K}$. Since the process is endothermic (heat is absorbed), the temperature of the solution decreases, so the final temperature is $22.0 \, °C - 4.3 \, °C = 17.7 \, °C$, which rounds to $17.9 \, °C$ based on the choices provided.

Commercial Cold Packs Consist of Solid Ammonium Nitrate and Water